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diff --git a/Chemistry/Chapter_16.ipynb b/Chemistry/Chapter_16.ipynb deleted file mode 100755 index 96bc5bbb..00000000 --- a/Chemistry/Chapter_16.ipynb +++ /dev/null @@ -1,584 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 16:Acid-Base Equilibria and Solubility Equilibria"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:16.1,Page no:715"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "InitCH3COOH1=0.2 #Initial concentration of CH3COOH solution, M\n",
- "#Let 'x' be the equilibrium concentration of the [H+] and [CH3COO-] ions after dissociation of [CH3COOH], M\n",
- "Ka=1.8*10**-5 #equilibrium constant of acid, M\n",
- "InitCH3COONa=0.3 #Initial concentration of CH3COONa solution and is equal to conc of Na+ and CH3COO- as it completely dissociates, M\n",
- "InitCH3COOH2=0.2 #Initial concentration of CH3COOH solution, M\n",
- "\n",
- "#Calculation\n",
- "#(a)\n",
- "import math\n",
- "x1=math.sqrt(Ka*InitCH3COOH1) #from the definition of ionisation constant Ka=[H+]*[CH3COO-]/[CH3COOH]=x*x/(0.2-x), which reduces to x*x/0.2, as x<<0.2 (approximation)\n",
- "pH1=-math.log10(x1) #since x is the conc. of [H+] ions\n",
- "\n",
- "#(b)\n",
- "#Let 'x' be the equilibrium concentration of the [H+] and hence conc of [CH3COO-] ions is '0.3 + x', M\n",
- "x2=Ka*InitCH3COOH2/InitCH3COONa #from the definition of ionisation constant Ka=[H+]*[CH3COO-]/[CH3COOH]=x*(0.3+x)/(0.2-x), which reduces to x*0.3/0.2(approximation)\n",
- "pH2=-math.log10(x2) #since x is the conc. of [H+] ions\n",
- "\n",
- "#Result\n",
- "print\"(a) the pH of CH3COOH solution is \",round(pH1,2)\n",
- "print\"(b) the pH of CH3COOH and CH3COONa solution is :\",round(pH2,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) the pH of CH3COOH solution is 2.72\n",
- "(b) the pH of CH3COOH and CH3COONa solution is : 4.92\n"
- ]
- }
- ],
- "prompt_number": 92
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.3,Page no:719"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "Ka=1.8*10**-5 #ionisation constant of acid\n",
- "InitCH3COONa=1 #Initial concentration of CH3COONa solution and is equal to conc of Na+ and CH3COO- as it completely dissociates, M\n",
- "InitCH3COOH=1 #Initial concentration of CH3COOH solution, M\n",
- "HCl=0.1 #moles of HCl added to 1L solution\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "#(a)\n",
- "x=Ka*InitCH3COOH/InitCH3COONa #from the definition of ionisation constant Ka=[H+]*[CH3COO-]/[CH3COOH]=x*(1+x)/(1-x), which reduces to x(approximation)\n",
- "pH=-math.log10(x) #since x is the conc. of [H+] ions\n",
- "#(b)\n",
- "CH3COO=InitCH3COONa-HCl #conc of CH3COO- ions, M\n",
- "CH3COOH=InitCH3COOH+HCl #conc of CH3COOH, M\n",
- "x2=Ka*CH3COOH/CH3COO #from the definition of ionisation constant Ka=[H+]*[CH3COO-]/[CH3COOH]=x*(0.9+x)/(1.1-x), which reduces to x*0.9/1.1(approximation)\n",
- "pH2=-math.log10(x2) #since x is the conc. of [H+] ions\n",
- "\n",
- "#Result\n",
- "print\"(a) the pH of CH3COOH and CH3COONa solution is :\",round(pH,2)\n",
- "print\"(b) the pH of solution after adding HCl is :\",round(pH2,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) the pH of CH3COOH and CH3COONa solution is : 4.74\n",
- "(b) the pH of solution after adding HCl is : 4.66\n"
- ]
- }
- ],
- "prompt_number": 93
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:16.5,Page no:728"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "InitCH3COOH=0.1 #Initial concentration of CH3COOH solution, M\n",
- "VCH3COOH=25 #volumeof CH3COOH, mL\n",
- "nCH3COOH=InitCH3COOH*VCH3COOH/1000 \n",
- "Ka=1.8*10**-5 #equilibrium constant of acid, M\n",
- "Kb=5.6*10**-10 #equilibrium constant of base, M\n",
- "N=0.1 #Initial concentration of NaOH solution, M\n",
- "V=10 #Initial volume of NaOH solution, mL\n",
- "\n",
- "\n",
- "#Calculation\n",
- "#(a)\n",
- "n=N*V/1000 #Initial moles of NaOH solution\n",
- "import math\n",
- "nCH3COOH_tit=nCH3COOH-n #moles of CH3COOH after titration\n",
- "nCH3COO=n #moles of CH3COO after titration\n",
- "H=nCH3COOH_tit*Ka/nCH3COO #conc of H+ ions, M\n",
- "pH=-math.log10(H) #since H is the conc. of [H+] ions\n",
- "\n",
- "#Result\n",
- "print\"(a) the pH of the solution is :\",round(pH,2)\n",
- "\n",
- "\n",
- "#Calculation\n",
- "#(b)\n",
- "V2=25.0 #Initial volume of NaOH solution, mL\n",
- "n2=N*V2/1000.0 #Initial moles of NaOH solution\n",
- "nCH3COOH_tit2=nCH3COOH-n2 #moles of CH3COOH after titration\n",
- "nCH3COO2=n2 #moles of CH3COO- ions after titration\n",
- "V_total=V2+VCH3COOH #total volume after titration\n",
- "CH3COO=nCH3COO2/V_total*1000 #conc of CH3COO- ions, M\n",
- "x=math.sqrt(Kb*CH3COO) #from the definition of ionisation constant Kb=[OH-]*[CH3COOH]/[CH3COO-]=x*x/(0.05-x), which reduces to x*x/0.05, as x<<0.05 (approximation)\n",
- "pOH=-math.log10(x) #since x is the conc. of [OH-] ions\n",
- "pH2=14.0-pOH \n",
- "\n",
- "#Result\n",
- "print\"(b) the pH of the solution is :\",round(pH2,2)\n",
- "\n",
- "#Calculation\n",
- "#(c)\n",
- "N=0.1 #Initial concentration of NaOH solution, M\n",
- "V=35 #Initial volume of NaOH solution, mL\n",
- "n=N*V/1000 #Initial moles of NaOH solution\n",
- "n_tit=n-nCH3COOH #moles of NaOH after titration\n",
- "nCH3COO=nCH3COOH #moles of CH3COO- ions after titration\n",
- "V_total=V+VCH3COOH #total volume\n",
- "OH=n_tit/V_total*1000 #conc of OH- ions, M\n",
- "pOH=-math.log10(OH) #since OH is the conc. of [OH-] ions\n",
- "pH=14-pOH \n",
- "\n",
- "#Result\n",
- "print\"(c) the pH of the solution is :\",round(pH,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a) the pH of the solution is : 4.57\n",
- "(b) the pH of the solution is : 8.72\n",
- "(c) the pH of the solution is : 12.22\n"
- ]
- }
- ],
- "prompt_number": 94
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:16.6,Page no:730"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "InitNH3=0.1 #Initial concentration of NH3 solution, M\n",
- "VNH3=25 #volume of NH3, mL\n",
- "nNH3=InitNH3*VNH3/1000 \n",
- "Ka=5.6*10**-10 #equilibrium constant of acid, M\n",
- "N=0.1 #Initial concentration, M\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "V=VNH3/InitNH3*N #Initial volume, mL\n",
- "V_total=V+VNH3 #total volume of the mixture, mL\n",
- "n_NH4Cl=nNH3 #moles of NH4Cl\n",
- "NH4Cl=n_NH4Cl/V_total*1000 #conc of NH4+ ions formed, M\n",
- "x=math.sqrt(Ka*NH4Cl) #from the definition of ionisation constant Ka=[H+]*[NH3]/[NH4+]=x*x/(NH4+-x), which reduces to x*x/NH4+, as x<<NH4+ (approximation)\n",
- "pH=-math.log10(x) #since x is the conc. of [H+] ions\n",
- "\n",
- "#Result\n",
- "print\"The pH of the solution at equivalent point is :\",round(pH,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The pH of the solution at equivalent point is : 5.28\n"
- ]
- }
- ],
- "prompt_number": 95
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example 16.8,Page no:738"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "solubility=0.67 #solubility of CaSO4, g/L\n",
- "\n",
- "#Calculation\n",
- "M=136.2 #mol mass of CaSO4, g\n",
- "s=solubility/M #concentration, M\n",
- "Ksp=s**2 #solubility product\n",
- "\n",
- "#Result\n",
- "print\"The Ksp of CaSO4 is :%.1e\"%Ksp"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The Ksp of CaSO4 is :2.4e-05\n"
- ]
- }
- ],
- "prompt_number": 96
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:16.9,Page no:739"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "Ksp=2.2*10**-20 #solubility product\n",
- "M=97.57 #mol mass of Cu(OH)2, g\n",
- "\n",
- "#Calculation\n",
- "s=(Ksp/4.0)**(1.0/3.0) #concentration, M\n",
- "solubility=s*M #solubility of Cu(OH)2, g/L\n",
- "\n",
- "#Result\n",
- "print\"The solubility of Cu(OH)2 is :%.2e\"%solubility,\"g/L\"\n",
- "\n",
- "#ALTERNATIVE METHOD:\n",
- "\n",
- "#Variable declaration\n",
- "Kspt=2.2*10**-20 #Ksp value from table 16.2\n",
- "M=97.75 #Molar mass of Cu(OH)2\n",
- "\n",
- "#Calculation\n",
- "from scipy.optimize import fsolve\n",
- "def f(s):\n",
- " Cu2=s\n",
- " OH_=2*s\n",
- " return(Kspt-(Cu2*(OH_**2)))\n",
- "s=fsolve(f,1)\n",
- "solubility=round(s,8)*M\n",
- "\n",
- "#Result\n",
- "print \"Alternative method: Solubility of Cu(OH)2=%.1e\"%solubility,\"g/L\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solubility of Cu(OH)2 is :1.72e-05 g/L\n",
- "Alternative method: Solubility of Cu(OH)2=1.8e-05"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " g/L\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:16.10,Page no:741"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "Ksp=1.1*10**-10 #solubility product of BaSO4\n",
- "#for Ba2+ ion\n",
- "N=0.004 #normality, M\n",
- "V=200 #vol in mL\n",
- "n=N*V/1000 #moles\n",
- "#for K2SO4sol\n",
- "N1=0.008 #normality, M\n",
- "V1=600 #vol in mL\n",
- "\n",
- "#Calculation\n",
- "n1=N1*V1/1000 #moles\n",
- "Nnew=n*1000/(V+V1) #conc of Ba2+ ions in final sol\n",
- "N1new=n1*1000/(V+V1) #conc of SO4 2- ions in final sol\n",
- "Q=Nnew*N1new #as Q=[Ba2+][SO4 2-]\n",
- "\n",
- "#Result\n",
- "print\"Q=\",Q\n",
- "if(Q>Ksp): #determination of precipitation\n",
- " print\"Q>Ksp, The solution is supersaturated and hence a precipitate will form\"\n",
- "else: \n",
- " print\"Q<Ksp, The solution is not supersaturated and hence a precipitate will not form\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Q= 6e-06\n",
- "Q>Ksp, The solution is supersaturated and hence a precipitate will form\n"
- ]
- }
- ],
- "prompt_number": 99
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:16.11,Page no:743"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "Br=0.02 #conc of Ag+ ions, M\n",
- "Ksp1=7.7*10**-13 #solubility product of AgBr\n",
- "Ksp2=1.6*10**-10 #solubility product of AgCl\n",
- "Cl=0.02 #conc of Cl- ions, M\n",
- "\n",
- "#Calculation\n",
- "#for Br\n",
- "Ag1=Ksp1/Br #conc of Ag+ ions at saturated state, M\n",
- "#for Cl\n",
- "Ag2=Ksp2/Cl #conc of Ag+ ions at saturated state, M\n",
- "\n",
- "#Result\n",
- "print \"[Ag+]=\",Ag2,\"M\"\n",
- "print\"To precipitate Br- without precipitating Cl- the concentration of Ag must be greater than %.1e\"%Ag1,\"M but less than\",Ag2,\"M\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "[Ag+]= 8e-09 M\n",
- "To precipitate Br- without precipitating Cl- the concentration of Ag must be greater than 3.9e-11 M but less than 8e-09 M\n"
- ]
- }
- ],
- "prompt_number": 100
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:16.12,Page no: 745"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "N_AgNO3=6.5*10**-3 #normality of AgNO3, M\n",
- "AgCl=143.4 #mol mass of AgCl, g\n",
- "Ksp=1.6*10**-10 #solubility product of AgCl\n",
- "\n",
- "#Calculation\n",
- "Ag=N_AgNO3 #conc of Ag+ ions as 's' is negligible, M\n",
- "s=Ksp/Ag #as Ksp=[Ag+][Cl-], molar solubility of AgCl, M\n",
- "solubility=s*AgCl #solubility of AgCl in AgBr solution, g/L\n",
- "\n",
- "#Result\n",
- "print\"The solubility of AgCl in AgBr solution is :%.2e\"%solubility,\"g/L\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The solubility of AgCl in AgBr solution is :3.53e-06 g/L\n"
- ]
- }
- ],
- "prompt_number": 101
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:16.14,Page no:748"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "FeCl2=0.003 #normality of FeCl2, M\n",
- "Fe=FeCl2 #as Fe2+ is strong electrolyte, conc of Fe2+=conc of FeCl2, M\n",
- "Ksp=1.6*10**-14 #solubility product of FeCl2\n",
- "Kb=1.8*10**-5 #ionisation constant of base\n",
- "\n",
- "#Calculation\n",
- "import math\n",
- "OH=math.sqrt(Ksp/Fe) #as Ksp=[Fe2+][OH-]**2, conc of OH- ions, M\n",
- "x=(OH**2)/Kb+OH #as Kb=[NH4+][OH-]/[NH3]\n",
- "\n",
- "#Result\n",
- "print\"To initiate precipitation the conc of NH3 must be slightly greater than :%.1e\"%x,\"M\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "To initiate precipitation the conc of NH3 must be slightly greater than :2.6e-06 M\n"
- ]
- }
- ],
- "prompt_number": 102
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:16.15,Page no:750"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "CuSO4=0.2 #normality of CuSO4, M\n",
- "NH3=1.2 #initial conc of NH3, M\n",
- "VNH3=1 #volume of NH3, L\n",
- "Kf=5*10**13 #formation constant\n",
- "\n",
- "#Calculation\n",
- "CuNH34=CuSO4 #conc of Cu(NH3)4 2+, M\n",
- "NH3=NH3-4*CuNH34 #conc of NH3 after formation of complex, as 4 moles of NH3 react to form 1 mole complex, M\n",
- "x=CuNH34/(NH3**4*Kf) #as Kf=[Cu(SO4)3 2+]/[Cu2+][NH3]**4\n",
- "\n",
- "#Result\n",
- "print\"The conc of Cu2+ ions in equilibrium is :%.1e\"%x,\"M\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The conc of Cu2+ ions in equilibrium is :1.6e-13 M\n"
- ]
- }
- ],
- "prompt_number": 103
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:16.16,Page no:751"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration\n",
- "InitNH3=1 #initial conc of NH3, M\n",
- "Ksp=1.6*10**-10 #solubility product of AgCl\n",
- "Kf=1.5*10**7 #formation constant of complex\n",
- "\n",
- "#Calculation\n",
- "K=Ksp*Kf #overall equilibrium constant\n",
- "import math\n",
- "s=math.sqrt(K)/(1+2*InitNH3*math.sqrt(K)) #molar solubility of AgCl, M\n",
- "\n",
- "#Result\n",
- "print\"Amount of AgCl which can be dissolved in 1 L of 1 M NH3 sol in equilibrium is :\",round(s,3),\"M\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Amount of AgCl which can be dissolved in 1 L of 1 M NH3 sol in equilibrium is : 0.045 M\n"
- ]
- }
- ],
- "prompt_number": 91
- }
- ],
- "metadata": {}
- }
- ]
-}
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