From f270f72badd9c61d48f290c3396004802841b9df Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:53:46 +0530 Subject: Removed duplicates --- .../Chapter8.ipynb | 258 +++++++++++++++++++++ 1 file changed, 258 insertions(+) create mode 100755 Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter8.ipynb (limited to 'Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter8.ipynb') diff --git a/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter8.ipynb b/Elements_of_Physical_Chemistry_by_Atkins_Peter/Chapter8.ipynb new file mode 100755 index 00000000..a3e364e2 --- /dev/null +++ b/Elements_of_Physical_Chemistry_by_Atkins_Peter/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 -- cgit