path: root/Textbook_of_Engineering_Chemistry/dChapter_2.ipynb

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 },
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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 2:Acids and Bases"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:1,Page no:38"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "OH=0.0025 #OH- concentration#\n",
      "K=1*10**-14#water ionization constant#\n",
      "\n",
      "#Calculation\n",
      "H=K/OH \n",
      "H=H/10**-12 \n",
      "\n",
      "#Result\n",
      "print\"The concentration of H+ ions is\",H*10**-12,\"M\" \n",
      " \n",
      "print\"\\nAs concentration of H+ is lesser than the concentration of OH-(0.0025) the cleaning solution will be basic in nature\" "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The concentration of H+ ions is 4e-12 M\n",
        "\n",
        "As concentration of H+ is lesser than the concentration of OH-(0.0025) the cleaning solution will be basic in nature\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2,Page no:40"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "pH=7.3 #pH value of human blood\n",
      "H=10**-pH \n",
      "\n",
      "#Calculation\n",
      "H1=H\n",
      "k=1*10**-14 #water ionization constant\n",
      "OH=k/H \n",
      "OH=OH\n",
      "\n",
      "#Result\n",
      "print\"H+ concentration of human blood is %.e\"%H1,\"M\" \n",
      "print\"\\nOH- concentration of human blood is %.3g\"%OH,\"M(in scientiifc form) or 0.2*10**-6 M\" "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "H+ concentration of human blood is 5e-08 M\n",
        "\n",
        "OH- concentration of human blood is 2e-07 M(in scientiifc form) or 0.2*10**-6 M\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:3,Page no:41"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#Variable declaration\n",
      "N1=0.2 #normality of HCl#\n",
      "V1=25 #volume of HCl in ml#\n",
      "M2=0.25 #molarity of NaOH#\n",
      "N2=M2*1 #normality of NaOH#\n",
      "V2=50 #volume of NaOH in ml#\n",
      "\n",
      "#Calculation\n",
      "V=V1+V2 #volume of resulting solution#\n",
      "N=(N2*V2-N1*V1)/V #normality of resulting solution#\n",
      "\n",
      "K=1*10**-14 #ionization constant of water#\n",
      "H=K/N \n",
      "H1=H/10**-13 \n",
      "\n",
      "pH=-math.log10(H) \n",
      "\n",
      "#Result\n",
      "\n",
      "print\"\\npH of the mixure will be\",pH"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "pH of the mixure will be 13.0\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:4,Page no:44"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "S=0.2 #salt concentration#\n",
      "A=0.2 #acid concentration#\n",
      "k=1.8*10**-5 #dissociation constant of acetic acid#\n",
      "\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "pH=-math.log10(k)+math.log10(S/A) \n",
      "v=1*10**-3 #amount of HCl added in lit#\n",
      "M=1 #molarity of HCl added#\n",
      "n=v*M #no of moles of HCl added per litre#\n",
      "A1=A+n \n",
      "S1=S-n \n",
      "pH2=-math.log10(k)+math.log10(S1/A1) \n",
      "p=pH-pH2 \n",
      "\n",
      "#Result\n",
      "print\"pH of the buffer solution before adding HCl is\",round(pH ,4)\n",
      "\n",
      "print\"\\npH of the buffer solution after adding HCl is\",round(pH2,3)\n",
      "print\"\\nAns: Change in pH is\",round(p,3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "pH of the buffer solution before adding HCl is 4.7447\n",
        "\n",
        "pH of the buffer solution after adding HCl is 4.74\n",
        "\n",
        "Ans: Change in pH is 0.004\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.1,Page no:46"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#(a)#\n",
      "#Variable declaration\n",
      "N1=1.0/1000.0 #normality of HCl#\n",
      "a=100.0 #percentage of ionization#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "C1=N1*a/100.0 \n",
      "pH1=-math.log10(C1) \n",
      "N2=1.0/10000.0 #normality of NaOH solution#\n",
      "C2=N2*a/10.0 \n",
      "C2a=C2/10.0**-4 \n",
      "k=10.0**-14 #dissociation constant of water#\n",
      "H2=k/C2 \n",
      "H2a=H2/10.0**-10 \n",
      "pH2=-math.log10(H2) \n",
      "N3=1.0/1000.0 #normality of NaOH solution#\n",
      "C3=N3*a/1000.0 \n",
      "C3a=C3/10.0**-3 \n",
      "H3=k/C3 \n",
      "H3a=H3/10.0**-11 \n",
      "pH3=-math.log10(H3) \n",
      "\n",
      "#Result\n",
      "print\"Ans(a)\\n(i)\\tThe pH of N/1000 HCl solution is\",pH1 \n",
      "print\"\\n(ii)\\tThe pH of the N/10000 solution is\",pH2 \n",
      "print\"\\n(iii)\\tThe pH of the N/1000 solution is\",pH3\n",
      "\n",
      "\n",
      "\n",
      "\n",
      "#(b)#\n",
      "#Variable declaration\n",
      "N=0.1 #normality of given weak base#\n",
      "pH=9.0 #pH of the base#\n",
      "H=10.0**(-pH) \n",
      "Ha=H/10.0**-9\n",
      "\n",
      "#Calculation\n",
      "OH=k/H \n",
      "OHa=OH/10.0**-5 \n",
      "a1=OH/N \n",
      "a1b=a1\n",
      "\n",
      "#Result\n",
      "print\"\\nAns(b)\\nDegree of ionization of given weak base is\",a1b,\"=\",a1b*100,\"%\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Ans(a)\n",
        "(i)\tThe pH of N/1000 HCl solution is 3.0\n",
        "\n",
        "(ii)\tThe pH of the N/10000 solution is 11.0\n",
        "\n",
        "(iii)\tThe pH of the N/1000 solution is 10.0\n",
        "\n",
        "Ans(b)\n",
        "Degree of ionization of given weak base is 0.0001 = 0.01 %\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.2,Page no:47"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "N=0.002 #normality of acetic acid solution#\n",
      "a=2.3 #percentage of ionization#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "H=N*a/100.0 #concentration of H+ ion#\n",
      "pH=-math.log10(H) \n",
      "\n",
      "#Result\n",
      "print\"\\nAns(a)\\n pH value of acid solution is\",round(pH,4)\n",
      "\n",
      "\n",
      "\n",
      "#Part b(b)#\n",
      "\n",
      "#Variable declaration\n",
      "N1=0.01 #normality of acetic acid solution#\n",
      "a1=60.0 #percentage of ionization#\n",
      "#ii#\n",
      "N2=0.1 #normality of acetic acid solution#\n",
      "a2=1.8 #percentage of ionization#\n",
      "#iii#\n",
      "N3=0.04 #normality of HNO3#\n",
      "a3=100.0 #percentage of ionization#\n",
      "#iv#\n",
      "W=4.0 #weight of NaOH dissolved in water in grams#\n",
      "EW=40.0 #equivalent weight weight of NaOH#\n",
      "\n",
      "\n",
      "#Calculation\n",
      "#i#\n",
      "H1=N1*a1/100.0 #concentration of H+ ion#\n",
      "pH1=-math.log10(H1) \n",
      "#ii#\n",
      "H2=N2*a2/100.0 #concentration of H+ ion#\n",
      "pH2=-math.log10(H2) \n",
      "#iii#\n",
      "H3=N3*a3/100.0 \n",
      "pH3=-math.log10(H3) \n",
      "N4=0.0001 #normality of Hcl#\n",
      "a4=100.0 #percentage of ionization#\n",
      "H4=N4*a4/100.0 \n",
      "pH4=-math.log10(H4) \n",
      "N5=1.0 #normality of Hcl#\n",
      "a5=100.0 #percentage of ionization#\n",
      "H5=N5*a5/100.0 \n",
      "pH5=-math.log10(H5) \n",
      "N6=0.1 #normality of HNO3#\n",
      "a6=100.0 #percentage of ionization#\n",
      "OH6=N6*a6/100.0 \n",
      "Kw=10.0**-14 \n",
      "H6=Kw/OH6 \n",
      "pH6=-math.log10(H6)\n",
      "N7=0.001 #normality of NaOH#\n",
      "a7=100.0 #percentage of ionization#\n",
      "OH7=N7*a7/100.0 \n",
      "Kw=10.0**-14 \n",
      "H7=Kw/OH7 \n",
      "pH7=-math.log10(H7) \n",
      "#iv#\n",
      "N8=W/EW \n",
      "a8=100.0 #percentage of ionization#\n",
      "OH8=N8*a8/100.0 \n",
      "Kw=10.0**-14 \n",
      "H8=Kw/OH8 \n",
      "pH8=-math.log10(H8) \n",
      "\n",
      "\n",
      "#Result\n",
      "print'\\nAns(b)\\n(i)   pH value of  0.01N acid solution is',round(pH1,4)\n",
      "\n",
      "print\"\\n(ii)  pH value of decinormal acid solution is\",round(pH2 ,4)\n",
      "\n",
      "print\"\\n(iii) The pH of 0.04N HNO3 solution is\",round(pH3,3)\n",
      "print\"\\n      The pH of 0.0001N Hcl solution is\",pH4 \n",
      "print\"\\n      The pH of 1N Hcl solution is\",pH5 \n",
      "print\"\\n      The pH of 0.1N NaOH solution is \",pH6\n",
      "print\"\\n      The pH of 0.01N NaOH solution is\",pH7\n",
      "\n",
      "print\"\\n(iv)  The pH of solution containing 4g NaoH solution is \",pH8 "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "Ans(a)\n",
        " pH value of acid solution is 4.3372\n",
        "\n",
        "Ans(b)\n",
        "(i)   pH value of  0.01N acid solution is 2.2218\n",
        "\n",
        "(ii)  pH value of decinormal acid solution is 2.7447\n",
        "\n",
        "(iii) The pH of 0.04N HNO3 solution is 1.398\n",
        "\n",
        "      The pH of 0.0001N Hcl solution is 4.0\n",
        "\n",
        "      The pH of 1N Hcl solution is -0.0\n",
        "\n",
        "      The pH of 0.1N NaOH solution is  13.0\n",
        "\n",
        "      The pH of 0.01N NaOH solution is 11.0\n",
        "\n",
        "(iv)  The pH of solution containing 4g NaoH solution is  13.0\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.3,Page no:48"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "N1=0.1 #normality of acetic acid#\n",
      "a1=1.3 #percentage of ionization#\n",
      "M1=10**-8 #molarity of hcl solution#\n",
      "a=100 #percentage of ionization#\n",
      "N2=0.05 #normality of Hcl#\n",
      "a2=100 #percentage of ionization#\n",
      "\n",
      "#Calculation\n",
      "#(a)#\n",
      "H1=N1*a1/100 \n",
      "#(b)#\n",
      "H=M1*a/100 \n",
      "pH=-math.log10(H) \n",
      "#(c)#\n",
      "pH2=-math.log10(N2*a2/100) \n",
      "M3=0.05 #molarity os H2SO4#\n",
      "a3=100 #percentage of ionization#\n",
      "pH3=-math.log10(M3*a3/100.0) \n",
      "\n",
      "#Result\n",
      "print\"(a).The hydrogen ion concentration of solution is %.2e\"%H1,\"g.ion/lit\"\n",
      "print'\\n(b).The pH of the Hcl solution is',pH\n",
      "print\"Theoretically the pH should be 8,however,the value will be close to 7 because H+ ions of water also plays a role\"\n",
      "print\"\\n(c).The pH of 0.05 Hcl solution is\",round(pH2,3)\n",
      "print\"The pH of 0.05M H2SO4 solution is\",round(pH3,3)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a).The hydrogen ion concentration of solution is 1.30e-03 g.ion/lit\n",
        "\n",
        "(b).The pH of the Hcl solution is 8.0\n",
        "Theoretically the pH should be 8,however,the value will be close to 7 because H+ ions of water also plays a role\n",
        "\n",
        "(c).The pH of 0.05 Hcl solution is 1.301\n",
        "The pH of 0.05M H2SO4 solution is 1.301\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.4,Page no:49"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "H1=0.005 #H+ ion concentration of solution in g.ion/lit#\n",
      "H2=3*10**-4 #H+ concentration of the solution#\n",
      "k=10**-14 #dissociation constant of water#\n",
      "OH3=0.1#hydroxyl concentration of a solution#\n",
      "k4=1.8*10**-5#dissociation constant of acetic acid at 180C#\n",
      "N4=0.1 #normality of acetic acid#\n",
      "N5=0.01 #normality of acetic acid#\n",
      "N6=0.001 #normality of acetic acid#\n",
      "\n",
      "#Calculation\n",
      "\n",
      "#Part-a#\n",
      "pH1=-math.log10(H1) \n",
      "\n",
      "#Part-b#\n",
      "pH2=-math.log10(H2) \n",
      "pOH2=14-pH2 \n",
      "OH2=k/H2\n",
      "\n",
      "#Part-c#\n",
      "H3=k/OH3 \n",
      "pH3=-math.log10(H3) \n",
      "V4=1/N4 \n",
      "\n",
      "#Part-d#\n",
      "a4=math.sqrt(k4*V4) #formula for degree of dissociation#\n",
      "H4=N4*a4 #H+ ion concentration#\n",
      "pH4=-math.log10(H4) \n",
      "V5=1/N5 \n",
      "a5=sqrt(k4*V5) #formula for degree of dissociation#\n",
      "H5=N5*a5 #H+ ion concentration#\n",
      "pH5=-math.log10(H5) \n",
      "V6=1/N6 \n",
      "a6=sqrt(k4*V6) #formula for degree of dissociation#\n",
      "H6=N6*a6 #H+ ion concentration#\n",
      "pH6=-math.log10(H6) \n",
      "\n",
      "\n",
      "#Result\n",
      "print\"\\n\\n(a)  The pH value of solution whose H+ ion concentration is 0.005g.ion/lit is\",round(pH1 ,3)\n",
      "print\"\\n(b)  The pH of a solution in which H+ is 3*10**-4 is\",round(pH2 ,2)\n",
      "print\"\\n     pOH of the solution is\",round(pOH2,2)\n",
      "print\"\\n     OH- concentration for a solution is%.1e\"%OH2,\"M\"\n",
      "print\"\\n(c)  pH of the solution whose hydroxyl concentration is N/10g.ion/lit is\",pH3\n",
      "print\"\\n(d)  pH of 0.1N acetic acid solution is\",round(pH4,3)\n",
      "print\"\\n     pH of 0.01N acetic acid solution is\",round(pH5,4)\n",
      "print\"\\n     pH of 0.001N acetic acid solution is\",round(pH6,4)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "(a)  The pH value of solution whose H+ ion concentration is 0.005g.ion/lit is 2.301\n",
        "\n",
        "(b)  The pH of a solution in which H+ is 3*10**-4 is 3.52\n",
        "\n",
        "     pOH of the solution is 10.48\n",
        "\n",
        "     OH- concentration for a solution is3.3e-11 M\n",
        "\n",
        "(c)  pH of the solution whose hydroxyl concentration is N/10g.ion/lit is 13.0\n",
        "\n",
        "(d)  pH of 0.1N acetic acid solution is 2.872\n",
        "\n",
        "     pH of 0.01N acetic acid solution is 3.3724\n",
        "\n",
        "     pH of 0.001N acetic acid solution is 3.8724\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.5,Page no:51"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "K1=10.0**-8 #dissociation constant of weak mono basic acid#\n",
      "N1=0.01 #normality of the acid#\n",
      "a2=4.0/100.0 #percentage of dissociation of acid at 20C#\n",
      "N2=0.1 #normality of acid#\n",
      "N3=0.1 #normality of HCl#\n",
      "N4=1.0/50.0 #normality of HCl#\n",
      "N5=0.01 #normality of H2SO4#\n",
      "\n",
      "#Calculation\n",
      "#Part-a#\n",
      "V1=1.0/N1 \n",
      "a1=math.sqrt(K1*V1) #degree of dissociation for weak acids#\n",
      "H1=N1*a1 #H+ concentration of the solution#\n",
      "pH1=-math.log10(H1) \n",
      "#Part-b#\n",
      "V2=1.0/N2 \n",
      "K2=(a2**2)/V2 \n",
      "#Part-c#\n",
      "pH3=-math.log10(N3) \n",
      "pH4=-math.log10(N4) \n",
      "pH5=-math.log10(N5) \n",
      "\n",
      "#Result\n",
      "print\"(a)  pH value of 0.01N solution of a weak mono basic acid is\",pH1 \n",
      "print\"\\n(b) The dissociation constant of the acid is  %.1e\"%K2\n",
      "print\"\\n(c) The pH of the 0.1N HCl solution is\",pH3\n",
      "print\"\\n    The pH of the 1/50N HCl solution is\",round(pH4,1)\n",
      "print\"\\n    The pH of the 0.01N H2SO4 solution is \",pH5"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)  pH value of 0.01N solution of a weak mono basic acid is 5.0\n",
        "\n",
        "(b) The dissociation constant of the acid is  1.6e-04\n",
        "\n",
        "(c) The pH of the 0.1N HCl solution is 1.0\n",
        "\n",
        "    The pH of the 1/50N HCl solution is 1.7\n",
        "\n",
        "    The pH of the 0.01N H2SO4 solution is  2.0\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.6,Page no:52"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "V1=50.0 #volume of Hcl in ml#\n",
      "V2=30.0 #volume of NaOH in ml#\n",
      "N1=1.0 #normality of Hcl#\n",
      "N2=1.0 #nomality of NaOH#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "V=V1+V2 #total volume of mixure of solutions#\n",
      "a=100.0 #percentage of ionization#\n",
      "N=(N1*V1-N2*V2)/V \n",
      "H=N*a/100 \n",
      "pH=-math.log10(H) \n",
      "\n",
      "#Result\n",
      "\n",
      "print'\\nThe pH of resultant solution is',round(pH,3)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The pH of resultant solution is 0.602\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.7,Page no:52"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "N1=1.0/10.0 #normality of NaOH#\n",
      "N2=1.0/20.0 #normality of HCl#\n",
      "V1=1.0 #volume of NaOH in lit#\n",
      "V2=1.0 #volume of HCl in lit#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "V=V1+V2 #volume of resultant solution#\n",
      "N=(N1*V1-N2*V2)/V \n",
      "k=1.0*10.0**-14 #ionization constant of water#\n",
      "H1=k/N \n",
      "H=H1/10.0**-13 \n",
      "pH=-math.log10(H1)\n",
      "\n",
      "#Result\n",
      "print\"\\npH of the solution is\",round(pH,1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "pH of the solution is 12.4\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.8,Page no:53"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "W=2.0 #weight of NaOH dissolved in water in grams#\n",
      "M=40.0 #molecular weight of NaOH#\n",
      "N=W/M #normality#\n",
      "a=100.0 #percentage of ionization#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "OH=N*a/100.0 #the OH- ion concentration of solution#\n",
      "Kw=10.0**-14 \n",
      "H=Kw/OH \n",
      "pH=-math.log10(H) \n",
      "\n",
      "#Result\n",
      "print'\\n The pH of the NaOH solution is',round(pH,1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        " The pH of the NaOH solution is 12.7\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.9,Page no:53"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "M=0.001 #molarity of benzoic acid#\n",
      "N=M #normality of benzoic acid#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "V=1/N \n",
      "K=7.3*10**-5 #dissociation constant of benzoic acid#\n",
      "a=math.sqrt(K*V) #since benzoic acid is very weak#\n",
      "\n",
      "#Result\n",
      "H=N*a \n",
      "print\"\\n The H+ concentration of the solution is%.3e\"%H,\"g.ion/litre\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        " The H+ concentration of the solution is2.702e-04 g.ion/litre\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.10,Page no:53"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "W=0.092 #weight of Formic acid per litre in grams#\n",
      "M=46 #molecular weight of Formic acid#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "N=W/M \n",
      "V=1/N \n",
      "K=2.4*10**-4 #Dissociation constant of Formic acid at 25C#\n",
      "a=math.sqrt(K*V) #For weak acids#\n",
      "\n",
      "#Result\n",
      "H=a*N \n",
      "print'\\n The H+ concentration of the solution is %.3e'%H"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        " The H+ concentration of the solution is 6.928e-04\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.11,Page no:54"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "k=2.5*10**-5 #dissociation constant of NH4OH#\n",
      "N=1.0/100.0 #normality of NH4OH#\n",
      "V=100\n",
      "#Calculation\n",
      "C=N #since volume of solution is one litre#\n",
      "NH=C \n",
      "NHOH=C \n",
      "OH1=k*NHOH/NH \n",
      "a=math.sqrt(k*V)\n",
      "\n",
      "#Result\n",
      "print\"Cong of OH- ions in solution is\",a*N,\"g ion per litre\"\n",
      "print\"\\nWhen 1/100 part of a g mol NH4Cl are dissolverd in a litre ,then ,\\nHydroxyl ion concentration in the solution is\",OH1\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Cong of OH- ions in solution is 0.0005 g ion per litre\n",
        "\n",
        "When 1/100 part of a g mol NH4Cl are dissolverd in a litre ,then ,\n",
        "Hydroxyl ion concentration in the solution is 2.5e-05\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.12,Page no:56"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "K=1.7*10**-5 #Dissociation constant of NH4OH#\n",
      "N=0.01 #Normality of NH4OH solution#\n",
      "\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "V=1.0/N \n",
      "a=math.sqrt(K*V) #since a is very small#\n",
      "OH=a*N \n",
      "\n",
      "NH4=0.05 #concentration of NH4+ in g.ion/lit#\n",
      "NH4OH=0.01 #concentration of NH4OH in g.mol/lit#\n",
      "OH2=K*NH4OH/NH4 \n",
      "#Result\n",
      "print\"\\nConcentration of OH- ions before addition of NH4Cl is %.2e\"%OH,\"g.ion/litre\"\n",
      "print\"\\nThe concentration of hydroxyl ions after adding NH4Cl is\",OH2,\"g.ion/litre\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "Concentration of OH- ions before addition of NH4Cl is 4.12e-04 g.ion/litre\n",
        "\n",
        "The concentration of hydroxyl ions after adding NH4Cl is 3.4e-06 g.ion/litre\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.13,Page no:56"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "k=1.8*10**-5 #dissociation constant of acetic acid at 18C#\n",
      "N=0.25 #normality of acetic acid solution#\n",
      "N2=0.25#normality os sodium acetate added#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "#Part-a#\n",
      "V=1/N \n",
      "a=math.sqrt(k*V) #formula of degree of dissociation for weak acids#\n",
      "H=N*a \n",
      "#Part-b#\n",
      "CH3COO=N2 \n",
      "CH3COOH=N2 \n",
      "H2=k*CH3COOH/CH3COO \n",
      "H3=H2/10**-5 \n",
      "a2=H2/N2 \n",
      "a3=a2/10**-5 \n",
      "\n",
      "\n",
      "#Result\n",
      "print\"(a)  0.25 N acetic acid solution---\"\n",
      "print\"\\tDegree of dissociation of acetic acid is %.3e\"%a\n",
      "print\"\\tH+ concentration of the solution is %.3e\"%H,\"g.ion/litre\"\n",
      "\n",
      "print\"\\n(b)  0.25 N acetic acid solution containing 0.25N sodium acetate----\"\n",
      "print\"\\tH+ ion concentration after adding sodium acetate is\",H3*10**-5\n",
      "print\"\\tDegree of dissociation after adding sodium acetate is\",a3*10**-5"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)  0.25 N acetic acid solution---\n",
        "\tDegree of dissociation of acetic acid is 8.485e-03\n",
        "\tH+ concentration of the solution is 2.121e-03 g.ion/litre\n",
        "\n",
        "(b)  0.25 N acetic acid solution containing 0.25N sodium acetate----\n",
        "\tH+ ion concentration after adding sodium acetate is 1.8e-05\n",
        "\tDegree of dissociation after adding sodium acetate is 7.2e-05\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.14,Page no:57"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "C1=0.06 #concentration od acetic acid in g.mol/lit#\n",
      "C2=0.04 #concentration of sodium acetate in g.mol/li#\n",
      "K=1.8*10**-5 #dissociation constant of acetic acid#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "H=K*C1/C2 \n",
      "pH=-math.log10(H) \n",
      "\n",
      "#Result\n",
      "print\"\\nThe pH of solution is\",round(pH,4)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The pH of solution is 4.5686\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.15,Page no:58"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "M1=0.2 #molarity of acetic acid#\n",
      "M2=0.2 #molarity of sodium acetate#\n",
      "K=1.8*10**-5 \n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "pH=-math.log10(K)+math.log10(M2/M1) #by using Henderson's equation#\n",
      "\n",
      "#Result\n",
      "print\"The pH value of buffer solution is\",round(pH,4)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The pH value of buffer solution is 4.7447\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.16,Page no:58"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "N=1.0/100.0 #normality of acetic acid#\n",
      "V=1.0/N \n",
      "k=1.8*10**-5 #dissociation constant of acetic acid#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "\n",
      "#Part-a#\n",
      "a=math.sqrt(k*V) #formula of degree of dissociation for weak acids#\n",
      "H=a*N \n",
      "\n",
      "#Part-b#\n",
      "n=0.01 #sodium acetate added in moles to one litre of acetic acid solution#\n",
      "CH3COO=n \n",
      "CH3COOH=n \n",
      "H1=k*CH3COOH/CH3COO \n",
      "\n",
      "\n",
      "#Result\n",
      "print\"(a)  H+ concentration of N/100 acetic acid solution is %.2e\"%H,\"g ion/litre\"\n",
      "print\"\\n(b)  H+ ion concentration in the solution after adding the sodium acetate is\",H1,\"g.ions/litre\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)  H+ concentration of N/100 acetic acid solution is 4.24e-04 g ion/litre\n",
        "\n",
        "(b)  H+ ion concentration in the solution after adding the sodium acetate is 1.8e-05 g.ions/litre\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.17,Page no:59"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "V=10 #volume of water in litres#\n",
      "N1=0.10 #moles of HCN added in solution#\n",
      "N2=0.10 #moles of NaCN added in solution#\n",
      "K=7.2*10**-10 #dissociation constant of HCN#\n",
      "CN=0.1 #CN- concentration#\n",
      "HCN=0.1 #HCN concentration#\n",
      "\n",
      "#Calculation\n",
      "H1=K*HCN/CN \n",
      "H=H1/10**-10 \n",
      "k=1*10**-14 #ionization constant of water#\n",
      "\n",
      "#Result\n",
      "print\"H+ concentration in the solution is\",H*10**-10\n",
      "OH=k/H1 \n",
      "print\"\\nOH- concentration in the solution is %.1e\"%OH"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "H+ concentration in the solution is 7.2e-10\n",
        "\n",
        "OH- concentration in the solution is 1.4e-05\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.18,Page no:59"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "K=1.7*10**-5 #dissociation constant of acid#\n",
      "pH=3.77#pH value of buffer solution#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "M=pH+math.log10(K) \n",
      "N=10**M #ratio of salt to acid#\n",
      "L=1/N\n",
      "\n",
      "#Result\n",
      "print\"The ratio of salt to acid in buffer is\",round(L)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The ratio of salt to acid in buffer is 10.0\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.19,Page no:59"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "k=1.8*10**-5 #dissociation constant of acetic acid#\n",
      "M=0.01 #molarity of acetic acid#\n",
      "N=M*1 #normality of acetic acid#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "V=1/N \n",
      "a=math.sqrt(k*V)#degree of dissociation for weak acids#\n",
      "H1=a/V \n",
      "H=H1/10**-4 \n",
      "pH=-math.log10(H1) \n",
      "\n",
      "#Result\n",
      "print\"Degree of dissociation of solution is %.2e\"%a \n",
      "print\"pH of the solution is\",round(pH ,4)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Degree of dissociation of solution is 4.24e-02\n",
        "pH of the solution is 3.3724\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.20,Page no:60"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "N1=0.2#concentration of acetic acid in g.molecule/lit#\n",
      "N2=0.25#concentration of sodium acetate in g.molecule/lit#\n",
      "K=1.8*10**-5#ionization constant of acetic acid at room temparature#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "pH1=-math.log10(K)+math.log10(N2/N1) \n",
      "N=1.0#normality of HCl added#\n",
      "V=0.5*10**-3#amount of HCl added in lit#\n",
      "M=N*V\n",
      "C1=N1+M#concentration of CH3COOH in moles/lit#\n",
      "C2=N2-M#concentration of CH3COONa in moles/lit#\n",
      "pH2=-math.log10(K)+math.log10(C2/C1)\n",
      "pH=pH1-pH2 \n",
      "\n",
      "#Result\n",
      "print\"pH value of the solution before adding HCl is\",round(pH1,4)\n",
      "print\"\\nThe pH of the solution after adding HCl is\",round(pH2,4)\n",
      "print\"\\nThe change of pH is\",round(pH,4)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "pH value of the solution before adding HCl is 4.8416\n",
        "\n",
        "The pH of the solution after adding HCl is 4.8397\n",
        "\n",
        "The change of pH is 0.002\n"
       ]
      }
     ],
     "prompt_number": 25
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.21,Page no:61"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "K=18*10**-6 #dissociation constant of NH4OH#\n",
      "N1=0.1 #normality of NH4OH solution#\n",
      "V=1.0/N1 \n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "a=math.sqrt(K*V)#since a is very small#\n",
      "OH=a/V \n",
      "W=2.0#weight of added NH4Cl in grams#\n",
      "M=53.0#molecular weight of NH4Cl#\n",
      "C=W/M \n",
      "C1=0.1 #concentration of NH4OH in g.mol/lit#\n",
      "OH2=K*C1/C\n",
      "CH3COO=0.02              #g ion per litre\n",
      "CH3COOH=0.2              #g mol per litre\n",
      "H_plus=K*CH3COOH/CH3COO\n",
      "pH=math.log10(H_plus)\n",
      "\n",
      "\n",
      "#Result\n",
      "print\"\\nThe concentration of hydroxyl ion before adding of NH4Cl is %.3e\"%OH,\"g ion per litre\"\n",
      "print\"\\nThe concentration of hydroxyl ion after adding 2g of NH4Cl is %.1e\"%OH2,\"g ion per litre\"\n",
      "print\"NOTE:Calculation mistake in book.(wrongly written as 48.6*10**-5,it should be 10**-6\"\n",
      "print \"\\npH is\",round(-pH,4)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "The concentration of hydroxyl ion before adding of NH4Cl is 1.342e-03 g ion per litre\n",
        "\n",
        "The concentration of hydroxyl ion after adding 2g of NH4Cl is 4.8e-05 g ion per litre\n",
        "NOTE:Calculation mistake in book.(wrongly written as 48.6*10**-5,it should be 10**-6\n",
        "\n",
        "pH is 3.7447\n"
       ]
      }
     ],
     "prompt_number": 27
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:2.22,Page no:62"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "ly=11.92 #equivalent conductvity of 0.02acetic acid solution in mho at 20C#\n",
      "lih=360 #the equivalent ionic conductance of an infinite dillution of hydrogen ion in mho#\n",
      "lic=40 #of acetate ion#\n",
      "li=lih+lic #of acetic acid#\n",
      "import math\n",
      "\n",
      "#Calculation\n",
      "a=ly/li #degree of dissociation#\n",
      "N=0.02 #normality of acetic acid#\n",
      "V=1/N \n",
      "K=(a**2)/V \n",
      "W=82 #mol.wt of CH3COONa#\n",
      "M=8.2#amount of sodium acetate added in g per litre solution#\n",
      "CH3COO=M/W \n",
      "H=K*N/CH3COO \n",
      "pH=-math.log10(H) \n",
      "\n",
      "#Result\n",
      "print\"Dissociation constant of acetic acid is %.2e\"%K\n",
      "print\"\\npH of the solution is\",round(pH,2)\n",
      "print\"\\nNOTE:\\n(i)Calculation  istake in calculation of K in book,exponent wrongly written as -6\"\n",
      "print\"(ii)pH is wrongly calculated in book as 3.45\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Dissociation constant of acetic acid is 1.78e-05\n",
        "\n",
        "pH of the solution is 5.45\n",
        "\n",
        "NOTE:\n",
        "(i)Calculation  istake in calculation of K in book,exponent wrongly written as -6\n",
        "(ii)pH is wrongly calculated in book as 3.45\n"
       ]
      }
     ],
     "prompt_number": 28
    }
   ],
   "metadata": {}
  }
 ]
}