{
"metadata": {
"name": "",
"signature": "sha256:f51aa40f01063be99829dfd06c53ed90029beb6946cceb4b92469b9c50e91012"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 14 - Development and use of activity concepts"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1 - pg 350"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the moles of Iodine present\n",
"#Initialization of variables\n",
"x1=0.0200\n",
"Kx=812.\n",
"#calculations\n",
"print \"Neglecting 2x in comparision with x1,\"\n",
"x=x1/Kx\n",
"#results\n",
"print '%s %.2e %s' %(\"Moles of Iodine present =\",x,\" mole\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Neglecting 2x in comparision with x1,\n",
"Moles of Iodine present = 2.46e-05 mole\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2 - pg 350 "
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Concentration of H+ ions\n",
"#Initialization of variables\n",
"Kc=1.749*10**-5 #M\n",
"n1=0.1 #mole\n",
"n2=0.01 #mole\n",
"#calculations\n",
"c=n1/n2 *Kc\n",
"#results\n",
"print '%s %.1e %s' %(\"Concentration of Hplus ions =\",c,\" M\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Concentration of Hplus ions = 1.7e-04 M\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3 - pg 351"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Concentraton of Hplus ions\n",
"#Initialization of variables\n",
"import math\n",
"c=0.01 #M\n",
"kc=1.749*10**-5 #M\n",
"#calculations\n",
"x2=c*kc\n",
"x=math.sqrt(x2)\n",
"#results\n",
"print '%s %.1e %s' %(\"Concentration of Hplus ions =\",x,\"M\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Concentration of Hplus ions = 4.2e-04 M\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4 - pg 351"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Concentration of OH- ions\n",
"#Initialization of variables\n",
"import math\n",
"K2=1.0008*10**-14 #m^2\n",
"K1=1.754*10**-5 #m\n",
"c=0.1\n",
"#calculations\n",
"print \"Neglecting x w.r.t c,\"\n",
"x2=c*K2/K1\n",
"x=math.sqrt(x2)\n",
"#results\n",
"print '%s %.1e %s' %(\"Concentration of OH minus ions =\",x,\" m\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Neglecting x w.r.t c,\n",
"Concentration of OH minus ions = 7.6e-06 m\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5 - pg 352"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Concentration of H plus ions\n",
"#Initialization of variables\n",
"import math\n",
"print \"from table 14.1,\"\n",
"r1=7.47*10**-5 #m\n",
"r2=4.57*10**-3 #m\n",
"mp=1.008*10**-14 #m**2\n",
"#calculations\n",
"r3=r2/r1\n",
"mH2=r3*mp\n",
"mH=math.sqrt(mH2)\n",
"#results\n",
"print '%s %.2e %s' %(\"Concentration of Hplus ions = \",mH,\" M\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"from table 14.1,\n",
"Concentration of Hplus ions = 7.85e-07 M\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6 - pg 354"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Concentraton of H+ ions\n",
"#Initialization of variables\n",
"print \"from table 14.1,\"\n",
"import math\n",
"r1=1.75*10**-5 #m\n",
"r2=1.772*10**-4 #m\n",
"mp=1.008*10**-14 #m**2\n",
"#calculations\n",
"r3=r2/r1\n",
"mH2=r3*mp\n",
"mH=math.sqrt(mH2)\n",
"#results\n",
"print '%s %.1e %s' %(\"Concentration of Hplus ions =\",mH,\" M\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"from table 14.1,\n",
"Concentration of Hplus ions = 3.2e-07 M\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 7 - pg 355"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Concentration of H+ ions\n",
"#Initialization of variables\n",
"import math\n",
"c=1*10**-6 #m\n",
"K=1.754*10**-5 #m\n",
"Kp=1.008*10**-14 #m**2\n",
"#calculations\n",
"mH=c\n",
"#Iteration 1\n",
"mOH=Kp/mH\n",
"mA=mH-mOH\n",
"mHA=mH*mA/K\n",
"mH2=mH-mHA+mOH\n",
"#Iteration 2\n",
"mOH2=Kp/mH2\n",
"mA2=mH2-mOH2\n",
"mHA2=mH2*mA2/K\n",
"mH3=mH2-mHA2+mOH2\n",
"#From x2\n",
"x2=math.sqrt(Kp)\n",
"x1=c\n",
"mOH3=Kp/x2\n",
"y2=x1\n",
"#From x1\n",
"mOH4=Kp/c\n",
"mA4=mH-mOH4\n",
"mHA4=mH*mA4/K\n",
"y1=c-mHA4-mA4\n",
"#upon further iterations, we get\n",
"mHplus=mH3\n",
"#results\n",
"print '%s %.2e %s' %(\"Concentration of H plus ions =\",mHplus,\"m\")\n",
"print 'The answer is a bit different due to rounding off error.'\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Concentration of H plus ions = 9.13e-07 m\n",
"The answer is a bit different due to rounding off error.\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8 - pg 358"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the values of dS0, dH0, Krm\n",
"#Initialization of variableH\n",
"print \"From table 14-3,\"\n",
"HH=0\n",
"HHcoo=-98.\n",
"HHcooh=-98.\n",
"SH=0\n",
"SHcoo=21.9\n",
"SHcooh=39.1\n",
"KH=0\n",
"KHcoo=58.64\n",
"KHcooh=62.38\n",
"#calculationH\n",
"dH=HH+HHcoo-HHcooh\n",
"dS=SH+SHcoo-SHcooh\n",
"dK=KH+KHcoo-KHcooh\n",
"K=10**dK\n",
"#results\n",
"print '%s %.1f %s' %(\" dS0 =\",dS,\"eu\")\n",
"print '%s %.1f %s' %(\"\\n dH0 =\",dH,\"kcal\")\n",
"print '%s %.2f' %(\"\\n log Krm =\",dK)\n",
"print '%s %.1e %s' %(\"\\n Krm =\",K,\"m\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From table 14-3,\n",
" dS0 = -17.2 eu\n",
"\n",
" dH0 = 0.0 kcal\n",
"\n",
" log Krm = -3.74\n",
"\n",
" Krm = 1.8e-04 m\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9 - pg 369"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Activity of cl and ca\n",
"#Initialization of variables\n",
"mca=0.01 #m\n",
"mcl=0.02 #m\n",
"#calculations\n",
"Mu=0.5*(mca*4 + mcl*1)\n",
"print \"From table 14-5,\"\n",
"aca=6 #A\n",
"acl=3 #A\n",
"print \"From table 14-6,\"\n",
"gaca=0.555 \n",
"gacl=0.843\n",
"Aca=gaca*mca\n",
"Acl=gacl*mcl\n",
"#results\n",
"print '%s %.4f' %(\"Activity of cl = \",Acl)\n",
"print '%s %.4f' %(\"\\n Activity of ca = \",Aca)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From table 14-5,\n",
"From table 14-6,\n",
"Activity of cl = 0.0169\n",
"\n",
" Activity of ca = 0.0056\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 10 - pg 369"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Concentration of H+ ions\n",
"#Initialization of variables\n",
"import math\n",
"m1=0.1 #m\n",
"m2=0.1 #m\n",
"K=1.754*10**-5 #m\n",
"#calculations\n",
"mu=0.5*(m1*1**2 + m2*1**2)\n",
"print(\"From table 14.5,\")\n",
"aH=9 #A\n",
"aA=4.5 #A\n",
"print(\"From table 14.6\")\n",
"gH=0.825\n",
"gA=0.775\n",
"gHA=1\n",
"x1=gHA*K/(gH*gA)\n",
"print(\"Assuming x to be small w.r.t m1,\")\n",
"x=math.sqrt(x1*m1)\n",
"#results\n",
"print '%s %.2e %s' %(\"Concentration of H plus ions =\",x,\" m\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From table 14.5,\n",
"From table 14.6\n",
"Assuming x to be small w.r.t m1,\n",
"Concentration of H plus ions = 1.66e-03 m\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11 - pg 372"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the concentration of H+ ions\n",
"#Initialization of variables\n",
"import math\n",
"import numpy\n",
"K=1.754*10**-5 #m\n",
"c=0.1\n",
"#calculations\n",
"print(\"Neglecting x w.r.t c,\")\n",
"x2=K\n",
"x=math.sqrt(K)\n",
"mu=x\n",
"print(\"From tables 14-5 and 14-6,\")\n",
"gH=0.963\n",
"gA=0.960\n",
"x22=K/(gH*gA)\n",
"p=([1,x22, -c*x22])\n",
"z=numpy.roots(p)\n",
"alpha=z[1]\n",
"#results\n",
"print '%s %.2e %s' %(\"concentration of H plus ions =\",alpha,\" m\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Neglecting x w.r.t c,\n",
"From tables 14-5 and 14-6,\n",
"concentration of H plus ions = 1.37e-03 m\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 12 - pg 373"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Solubility of Agcl\n",
"#Initialization of variables\n",
"print(\"From table 14.3\")\n",
"import math\n",
"K1=-13.5089\n",
"K2=-22.9792\n",
"K3=19.2218\n",
"c=0.1 #m\n",
"#calculations\n",
"logK=K1-K2-K3\n",
"K=10**logK\n",
"mu=0.5*(c*1**2 + c*1**2)\n",
"print(\"From tables 14-5 and 14-6,\")\n",
"gAg=0.745\n",
"gCl=0.755\n",
"x2=K/(gAg*gCl)\n",
"x=math.sqrt(x2)\n",
"#results\n",
"print '%s %.2e %s' %(\"Solubility of Agcl =\",x,\"m\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"From table 14.3\n",
"From tables 14-5 and 14-6,\n",
"Solubility of Agcl = 1.78e-05 m\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 13 - pg 376"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the Concentration of Na and Cl in both cases\n",
"#Initialization of variables\n",
"import numpy\n",
"Cna=0.11\n",
"Ccl=0.1\n",
"#calculations\n",
"p=([99, - 2.1, Cna*Ccl])\n",
"z=numpy.roots(p)\n",
"alpha=z[1]\n",
"Na1=Cna-10*alpha\n",
"Cl1=Ccl-10*alpha\n",
"#results\n",
"print '%s %.4f %s' %(\" Concentration of Na in 1 =\",Na1,\"M\")\n",
"print '%s %.4f %s' %(\"\\n Concentration of Cl in 1 =\",Cl1,\" M\")\n",
"print '%s %.4f %s' %(\"\\n Concentration of Na in 2 =\",alpha,\"M\")\n",
"print '%s %.4f %s' %(\"\\n Concentration of Cl in 2 =\",alpha,\"M\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Concentration of Na in 1 = 0.0157 M\n",
"\n",
" Concentration of Cl in 1 = 0.0057 M\n",
"\n",
" Concentration of Na in 2 = 0.0094 M\n",
"\n",
" Concentration of Cl in 2 = 0.0094 M\n"
]
}
],
"prompt_number": 14
}
],
"metadata": {}
}
]
}