From 6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:53:46 +0530 Subject: Removed duplicates --- Chemistry_by_Raymond_Chang/Chapter_15.ipynb | 472 ++++++++++++++++++++++++++++ 1 file changed, 472 insertions(+) create mode 100755 Chemistry_by_Raymond_Chang/Chapter_15.ipynb (limited to 'Chemistry_by_Raymond_Chang/Chapter_15.ipynb') diff --git a/Chemistry_by_Raymond_Chang/Chapter_15.ipynb b/Chemistry_by_Raymond_Chang/Chapter_15.ipynb new file mode 100755 index 00000000..fe8c12e3 --- /dev/null +++ b/Chemistry_by_Raymond_Chang/Chapter_15.ipynb @@ -0,0 +1,472 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 15:Acids and Bases" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:15.2,Page no:662" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "OH=0.0025 # [OH-] ion concentration, M\n", + "Kw=1*10**-14 # ionic product of water, M**2\n", + "#Calculation\n", + "H=Kw/OH # From the formula (ionic product)Kw=[H+]*[OH-]\n", + "#Result\n", + "print\"The [H+] ion concentration of the solution is :\",H,\"M\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The [H+] ion concentration of the solution is : 4e-12 M\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:15.3,Page no:665" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "H1=3.2*10**-4 #Concentration of [H+] ion on first occasion,\n", + "H2=1*10**-3 #Concentration of [H+] ion on second occasion, M\n", + "\n", + "#Calculation\n", + "pH1=-math.log10(H1) #from the definition of pH\n", + "pH2=-math.log10(H2) #from the definition of pH\n", + "\n", + "#Result\n", + "print\"pH of the solution on first occasion is:\",round(pH1,2)\n", + "print\"pH of the solution on second occasion is :\",pH2\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "pH of the solution on first occasion is: 3.49\n", + "pH of the solution on second occasion is : 3.0\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:15.4,Page no:665" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "pH=4.82 #Given\n", + "\n", + "#Calculation\n", + "H=10**(-pH) #Concentration of [H+] ion, M, formula from the definition of pH\n", + "\n", + "#Result\n", + "print\"The [H+] ion concentration of the solution is :%.1e\"%H,\"M\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The [H+] ion concentration of the solution is :1.5e-05 M\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:15.5,Page no:666" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#Variable declaration\n", + "OH=2.9*10**-4 #Concentration of [OH-] ion, M\n", + "#Calculation\n", + "pOH=-math.log10(OH) #by definition of p(OH)\n", + "pH=14-pOH \n", + "#Result\n", + "print\"\\t the pH of the solution is :\",round(pH,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\t the pH of the solution is : 10.46\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:15.6,Page no:669" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "ConcHCl=1*10**-3 #Concentration of HCl solution, M\n", + "ConcBaOH2=0.02 #Concentration of Ba(OH)2 solution, M\n", + "\n", + "#Calculation\n", + "#for HCL solution\n", + "H=ConcHCl #Concentration of [H+] ion after ionisation of HCl\n", + "pH=-math.log10(H) \n", + "#for Ba(OH)2 solution\n", + "OH=ConcBaOH2*2 #Concentration of [OH-] ion after ionisation of Ba(OH)2 as two ions are generated per one molecule of Ba(OH)2\n", + "pOH=-math.log10(OH) \n", + "pH2=14-pOH \n", + "\n", + "#Result\n", + "print\"(a). The pH of the HCl solution is :\",pH\n", + "print\"(b). The pH of the Ba(OH)2 solution is :\",round(pH2,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a). The pH of the HCl solution is : 3.0\n", + "(b). The pH of the Ba(OH)2 solution is : 12.6\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:15.8,Page no:675" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#Variable declaration\n", + "InitHNO2=0.036 #Initial concentration of HNO2 solution, M\n", + "#Let 'x' be the equilibrium concentration of the [H+] and [NO2-] ions, M\n", + "Ka=4.5*10**-4 #ionisation constant of HNO2, M\n", + "\n", + "#Calculation\n", + "x=math.sqrt(Ka*InitHNO2) #from the definition of ionisation constant Ka=[H+]*[NO2-]/[HNO2]=x*x/(0.036-x), which reduces to x*x/0.036, as x<5):\n", + " x1=(-Ka+math.sqrt((Ka**2)-(-4*1*Ka*InitHNO2)))/(2*1) \n", + " x2=(-Ka-math.sqrt((Ka**2)-(-4*1*Ka*InitHNO2)))/(2*1) \n", + " if(x1>0):\n", + " x=x1 \n", + " else :\n", + " x=x2 \n", + "pH=-math.log10(x) #since x is the conc. of [H+] ions\n", + "\n", + "#Result\n", + "print\"The pH of the HNO2 solution is :\",round(pH,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pH of the HNO2 solution is : 2.42\n" + ] + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:15.9,Page no:676" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "pH=2.39 # pH of the HCOOH acid solution\n", + "InitHCOOH=0.1 #initial concentration of the solution\n", + "\n", + "#Calculation\n", + "H=10**(-pH) #[H+] ion concentration from the definition of pH, M\n", + "HCOO_=H #[HCOO-] ion concentration,M\n", + "HCOOH=InitHCOOH-H #HCOOH concentration in M\n", + "Ka=(H*HCOO_)/(HCOOH) #ionisation constant of the acid, M, Ka=[H+]*[HCOO-]/[HCOOH]\n", + "\n", + "#Result\n", + "print\"The ionisation constant of the given solution is :%.2e\"%Ka,\"M(approx)\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ionisation constant of the given solution is :1.73e-04 M(approx)\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:15.10,Page no:678" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "InitNH3=0.4 #Initial concentration of NH3 solution, M\n", + "#Let 'x' be the equilibrium concentration of the [OH-] and [NH4+] ions, M\n", + "Kb=1.8*10**-5 #ionisation constant of NH3, M\n", + "\n", + "#Calculation\n", + "x=math.sqrt(Kb*InitNH3) #from the definition of ionisation constant Kb=[OH-]*[NH4+]/[NH3]=x*x/(InitNH3-x), which reduces to x*x/InitNH3, as x<5):\n", + " x1=(-Kb+math.sqrt((Kb**2)-(-4*1*Kb*InitNH3)))/(2*1) \n", + " x2=(-Kb-math.sqrt((Kb**2)-(-4*1*Kb*InitNH3)))/(2*1) \n", + " if(x1>0):#as only one root is positive\n", + " x=x1 \n", + " else:\n", + " x=x2 \n", + "pOH=-math.log10(x) #since x is the conc. of [H+] ions\n", + "pH=14-pOH \n", + "\n", + "#Result\n", + "print\"The pH of the NH3 solution is :\",round(pH,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pH of the NH3 solution is : 11.43\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:15.11,Page no:682" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "InitC2H2O4=0.1 #Initial concentration of C2H2O4 solution, M\n", + "#Let 'x1' be the equilibrium concentration of the [H+] and [C2HO4-] ions, M\n", + "#First stage of ionisation\n", + "import math\n", + "Kw=10**-14 #ionic product of water, M**2\n", + "Ka1=6.5*10**-2 #ionisation constant of C2H2O4, M\n", + "\n", + "#Calculation\n", + "x=math.sqrt(Ka1*InitC2H2O4) #from the definition of ionisation constant Ka1=[H+]*[C2HO4-]/[C2H2O4]=x*x/(InitC2H2O4-x), which reduces to x*x/InitC2H2O4, as x<5):\n", + " x1=(-Ka1+math.sqrt((Ka1**2)-(-4*1*Ka1*InitC2H2O4)))/(2*1) \n", + " x2=(-Ka1-math.sqrt((Ka1**2)-(-4*1*Ka1*InitC2H2O4)))/(2*1) \n", + " if(x1>0):#as only one root is positive\n", + " x=x1 \n", + " else: \n", + " x=x2 \n", + "C2H2O4=InitC2H2O4-x #equilibrium value\n", + "\n", + "#Result\n", + "print\"The concentration of the [H2C2O4] in the solution is :\",round(C2H2O4,3),\"M\"\n", + "\n", + "\n", + "#Second stage of ionisation\n", + "\n", + "#Variable declaration\n", + "InitC2HO4=x #concentration of C2HO4 from first stage of ionisation\n", + "Ka2=6.1*10**-5 #ionisation constant of C2HO4-, M\n", + "\n", + "#Calculation\n", + "#Let 'y' be the concentration of the [C2HO4-] dissociated to form [H+] and [C2HO4-] ions, M\n", + "y=Ka2 #from the definition of ionisation constant Ka2=[H+]*[C2O4-2]/[C2HO4-]=(0.054+y)*y/(0.054-y), which reduces to y, as y<5):\n", + " x1=(-Ka2+math.sqrt((Ka2**2)-(-4*1*Ka2*InitC2HO4)))/(2*1) \n", + " x2=(-Ka2-math.sqrt((Ka2**2)-(-4*1*Ka2*InitC2HO4)))/(2*1) \n", + " if(x1>0):#as only one root is positive\n", + " y=x1 \n", + " else: \n", + " y=x2 \n", + "C2HO4=InitC2HO4-y #from first and second stages of ionisation\n", + "H=x+y #from first and second stages of ionisation\n", + "C2O4=y #from the assumption\n", + "OH=Kw/H # From the formula (ionic product)Kw=[H+]*[OH-]\n", + "\n", + "#Result\n", + "print\"The concentration of the [HC2O4-] ion in the solution is :\",round(C2HO4,3),\"M\"\n", + "print\"The concentration of the [H+] ion in the solution is :\",round(H,3),\"M\" \n", + "print\"The concentration of the [C2O4^2-] ion in the solution is :\",C2O4,\"M\"\n", + "print\"The concentration of the [OH-] ion in the solution is :%.1e\"%OH,\"M\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The concentration of the [H2C2O4] in the solution is : 0.046 M\n", + "The concentration of the [HC2O4-] ion in the solution is : 0.054 M\n", + "The concentration of the [H+] ion in the solution is : 0.054 M\n", + "The concentration of the [C2O4^2-] ion in the solution is : 6.1e-05 M\n", + "The concentration of the [OH-] ion in the solution is :1.8e-13 M\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example no:15.13,Page no:690" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "InitCH3COONa=0.15 #Initial concentration of CH3COONa solution, M\n", + "InitCH3COO=InitCH3COONa #concentration of [CH3COO-] ion after dissociation of CH3COONa solution, M\n", + "#Let 'x' be the equilibrium concentration of the [OH-] and [CH3COOH] ions after hydrolysis of [CH3COO-], M\n", + "Kb=5.6*10**-10 #equilibrium constant of hydrolysis, M\n", + "import math\n", + "\n", + "#Calculation\n", + "x=math.sqrt(Kb*InitCH3COO) #from the definition of ionisation constant Kb=[OH-]*[CH3COOH]/[CH3COO-]=x*x/(0.15-x), which reduces to x*x/0.15, as x<<0.15 (approximation)\n", + "approx=x/InitCH3COO*100 #this is the percentage of approximation taken. if it is more than 5%, we will be having higher deviation from correct value\n", + "if(approx>5):\n", + " x1=(-Kb+math.sqrt((Kb**2)-(-4*1*Kb*InitCH3COO)))/(2*1) \n", + " x2=(-Kb-math.sqrt((Kb**2)-(-4*1*Kb*InitCH3COO)))/(2*1) \n", + " if(x1>0):#as only one root is positive\n", + " x=x1 \n", + " else: \n", + " x=x2 \n", + "pOH=-math.log10(x) #since x is the conc. of [OH-] ions\n", + "pH=14-pOH \n", + "\n", + "#Result\n", + "print\"The pH of the salt solution is :\",round(pH,2)\n", + "percenthydrolysis=x/InitCH3COO*100 \n", + "print\"The percentage of hydrolysis of the salt solution is :\",round(percenthydrolysis,4),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pH of the salt solution is : 8.96\n", + "The percentage of hydrolysis of the salt solution is : 0.0061 %\n" + ] + } + ], + "prompt_number": 51 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit