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| author | hardythe1 | 2015-01-28 14:31:21 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-01-28 14:31:21 +0530 |
| commit | 9add422993fb2649287260bc91d429a07d1810d5 (patch) | |
| tree | ef48c2b2579e65b982d3f700c4fa76b81d2496c1 /Elements_of_Physical_Chemistry/Chapter8.ipynb | |
| parent | 6e3407ba85ae84e1cee1ae0c972fd32c5504d827 (diff) | |
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diff --git a/Elements_of_Physical_Chemistry/Chapter8.ipynb b/Elements_of_Physical_Chemistry/Chapter8.ipynb new file mode 100755 index 00000000..a3e364e2 --- /dev/null +++ b/Elements_of_Physical_Chemistry/Chapter8.ipynb @@ -0,0 +1,258 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8 - Consequences of equilibrium"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I1 - Pg 176"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the equlibrium pH\n",
+ "#Initialization of variables\n",
+ "ph1=6.37\n",
+ "ph2=10.25\n",
+ "ph3=7.21\n",
+ "ph4=12.67\n",
+ "#calculations\n",
+ "pH1=0.5*(ph1+ph2)\n",
+ "pH2=0.5*(ph3+ph4)\n",
+ "#results\n",
+ "print '%s %.2f' %(\"Equilibrium pH in case 1 = \",pH1)\n",
+ "print '%s %.2f' %(\"\\n Equilibrium pH in case 2 = \",pH2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equilibrium pH in case 1 = 8.31\n",
+ "\n",
+ " Equilibrium pH in case 2 = 9.94\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example I2 - Pg 178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the pH of the solution\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "n=2.5/1000. #mol\n",
+ "C=0.2 #mol/L\n",
+ "vbase=37.5/1000. #L\n",
+ "#calculations\n",
+ "V=n/C\n",
+ "base=n/vbase\n",
+ "H=math.pow(10,-14) /base\n",
+ "print '%s' %(\"It follows from example 8.2 that\")\n",
+ "pH=10.2\n",
+ "#results\n",
+ "print '%s %.1f' %(\"\\n pH of the solution = \",pH)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "It follows from example 8.2 that\n",
+ "\n",
+ " pH of the solution = 10.2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the percent of acetic acid molecules that have donated a proton\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "C=0.15 #M\n",
+ "Ka=1.8*math.pow(10,-5)\n",
+ "#calculations\n",
+ "x=math.sqrt(C*Ka)\n",
+ "f=x/C\n",
+ "percent=f*100.\n",
+ "#results\n",
+ "print '%s %.1f %s' %(\"percent of acetic acid molecules that have donated a proton =\",percent,\"percent\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "percent of acetic acid molecules that have donated a proton = 1.1 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 171"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the fraction proportionated\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "pKa=4.88\n",
+ "C=0.01 #M\n",
+ "pKw=14\n",
+ "#calculations\n",
+ "pKb=pKw-pKa\n",
+ "Kb=math.pow(10,(-pKb))\n",
+ "x=(math.sqrt(C*Kb))\n",
+ "pOH=-math.log(x)\n",
+ "pH=14-pOH\n",
+ "f=x/C\n",
+ "#results\n",
+ "print '%s %.1e' %(\"fraction protonated = \",f)\n",
+ "print '%s %d' %(\"\\n 1 molecule in about \",1./f)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "fraction protonated = 2.8e-04\n",
+ "\n",
+ " 1 molecule in about 3630\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate te concentration of carbonate ions\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "pKa2=10.25\n",
+ "#calculations\n",
+ "C=math.pow(10,(-pKa2))\n",
+ "#results\n",
+ "print '%s %.1e %s' %(\"Concentration of Carbonate ions =\",C,\" mol/l\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Concentration of Carbonate ions = 5.6e-11 mol/l\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 - Pg 173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the final pH of the solution\n",
+ "#Initialization of variables\n",
+ "import math\n",
+ "vOH=5*math.pow(10,-3) #L\n",
+ "vHClO=25*math.pow(10,-3) #L\n",
+ "C=0.2 #mol/L\n",
+ "#calculations\n",
+ "nOH=vOH*C\n",
+ "nHClO=vHClO*C/2.\n",
+ "nrem=nHClO-nOH\n",
+ "pH=7.53-math.log10(nrem/nOH)\n",
+ "#results\n",
+ "print '%s %.1f' %(\"Final pH= \",pH)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Final pH= 7.4\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |