updated the code

20 files changed, 100 insertions, 92 deletions

diff --git a/3428/CH23/EX14.23.1/Ex14_23_1.sce b/3428/CH23/EX14.23.1/Ex14_23_1.sce
index 19dfb324d..b1310dd1d 100644
--- a/3428/CH23/EX14.23.1/Ex14_23_1.sce
+++ b/3428/CH23/EX14.23.1/Ex14_23_1.sce
@@ -3,10 +3,10 @@
 clc;

 //K_p=K_c*((R*T)^dl_n)

 dl_n=2-(1+3)

-T=673

-R=0.0821

+T=673                  //Kelvin

+R=0.0821              // (dm^3K^-1mol^-1)

 K_c=0.495

-P=2

+P=2                    //atm

 K_p=K_c*((R*T)^dl_n) 

 disp(K_p)

 //K_p=K_x*((P)^dl_n)

diff --git a/3428/CH23/EX14.23.11/Ex14_23_11.sce b/3428/CH23/EX14.23.11/Ex14_23_11.sce
index 8d68f0791..a6f0210aa 100644
--- a/3428/CH23/EX14.23.11/Ex14_23_11.sce
+++ b/3428/CH23/EX14.23.11/Ex14_23_11.sce
@@ -1,7 +1,8 @@
 //Section-14,Example-6,Page no.-PC.84

-//To find the concentration of Ag2+ for the given conditions.

-clc;

-K_spAgI=1.5*10^-16

-K_spAgCl=1.56*10^-10                       //C=[I-]/[Cl-]

-C=(K_spAgI)/(K_spAgCl)

-disp(C,'Required concentration of AgCl')

+//To find the concentration of Ag+ ions at which I- and Cl- ions will precipitate.

+clc,

+Ksp_AgI=1.5*10^-16

+Ksp_AgCl=1.56*10^-10

+C=Ksp_AgI/Ksp_AgCl

+disp(C,'concentration of Ag+ ions at which I- and Cl- ions will precipitate')

+//Answer in the book is  9.6*10^-7 approximately equal to 0.0000010.

diff --git a/3428/CH23/EX14.23.12/Ex14_23_12.sce b/3428/CH23/EX14.23.12/Ex14_23_12.sce
deleted file mode 100644
index 2e4cf103f..000000000
--- a/3428/CH23/EX14.23.12/Ex14_23_12.sce
+++ /dev/null
@@ -1,8 +0,0 @@
-//Section-14,Example-6,Page no.-PC.84

-//To calculate the free energy change and justify the given reactions.

-clc;

-//Cu_2S + O_2 = 2Cu + SO_2

-dl_G1=88.2

-dl_G2=300.1

-dl_G=dl_G1-dl_G2

-disp(dl_G,'Free energy change(kJ)')

diff --git a/3428/CH23/EX14.23.13/Ex14_23_13.sce b/3428/CH23/EX14.23.13/Ex14_23_13.sce
index 73e257841..d35067ad8 100644
--- a/3428/CH23/EX14.23.13/Ex14_23_13.sce
+++ b/3428/CH23/EX14.23.13/Ex14_23_13.sce
@@ -1,7 +1,8 @@
-//Section-14,Example-1,Page no.-PC.90
-//To justify the given reaction
-clc,
-dl_G1=88.2
-dl_G2=-300.1
-dl_G=dl_G1+dl_G2
-disp(dl_G)
+//Section-14,Example-1,Page no.-PC.90

+//To calculate the free energy change and justify the given reactions.

+clc;

+//Cu_2S + O_2 = 2Cu + SO_2

+dl_G1=88.2            //kJ

+dl_G2=300.1            //kJ

+dl_G=dl_G1-dl_G2         //kJ

+disp(dl_G,'Free energy change(kJ)')

diff --git a/3428/CH23/EX14.23.15/Ex14_23_15.sce b/3428/CH23/EX14.23.15/Ex14_23_15.sce
index f60b85e99..222d348ae 100644
--- a/3428/CH23/EX14.23.15/Ex14_23_15.sce
+++ b/3428/CH23/EX14.23.15/Ex14_23_15.sce
@@ -2,6 +2,6 @@
 //To estimate fraction of Morphine protonated.

 clc;

 K_b=1.6*10^-6

-B=0.010

+B=0.010                    //mol/L

 F_pro=sqrt(K_b/B)

 disp(F_pro,'Fraction of Morphine protonated')

diff --git a/3428/CH23/EX14.23.16/Ex14_23_16.sce b/3428/CH23/EX14.23.16/Ex14_23_16.sce
index 7156a46a1..e46413cfe 100644
--- a/3428/CH23/EX14.23.16/Ex14_23_16.sce
+++ b/3428/CH23/EX14.23.16/Ex14_23_16.sce
@@ -1,8 +1,9 @@
 //Section-14,Example-2,Page no.-PC.111

 //To estimate the pH of 0.10 M NH_3 at 25 degree Celcius.

-clc;

+clc

+T=25                  //degree Celsius

 K_b=1.8*10^-5

-B=0.1

+B=0.1                //M

 pK_b=-log10(K_b)

 pK_w=14

 F_pro=sqrt(K_b/B)                      //Fraction protonated

diff --git a/3428/CH23/EX14.23.17/Ex14_23_17.sce b/3428/CH23/EX14.23.17/Ex14_23_17.sce
index 97bdadbf2..9f8d59de8 100644
--- a/3428/CH23/EX14.23.17/Ex14_23_17.sce
+++ b/3428/CH23/EX14.23.17/Ex14_23_17.sce
@@ -1,22 +1,24 @@
 //Section-14,Example-1,Page no.-PC.112
 //To calculate the pH values of the following.
 clc;
-C_HCl=0.001
-C_1=C_HCl               //Since HCl is a strong acid,[H3O+]=[HCl]
+C_HCl=0.001                 //(M)
+C_1=C_HCl               //Since HCl is a strong acid,[H3O+]=[HCl](M)
 pH_1=-log10(C_1)
 disp(pH_1,'pH of 0.001 M HCl')
-C_NaOH=0.0001
-C_2=C_NaOH            //Since NaOH is a strong base so [OH-]=[NaOH]
+C_NaOH=0.0001             //(M)
+C_2=C_NaOH            //Since NaOH is a strong base so [OH-]=[NaOH](M)
 pOH=-log10(C_2)
 pH_2=14-pOH
 disp(pH_2,'pH of 0.0001 M NaOH')
-C_BaOH2=0.001
-C_31=2*C_BaOH2            // [OH-]=2*[Ba(OH)2]
+C_BaOH2=0.001              //(M)
+C_31=2*C_BaOH2            // [OH-]=2*[Ba(OH)2](M)
 k_w=10^-14
-C_32=k_w/(C_31)
+C_32=k_w/(C_31)         //(M)
 pH_3=-log10(C_32)
 disp(pH_3,'pH of 0.001 M Ba(OH)2')
-M=(0.049/98)/(200/1000)                //Molarity of H_2SO_4 solution
-C_4=2*M                               //[H_3O+]
+M_H2SO4=0.0049                                 //Mass of H2SO4 (gm)
+V_req=200                           //Volume of solution to be prepared(ml)
+Mo=(M_H2SO4/98)/(V_req/1000)                //Molarity of H_2SO_4 solution(M)
+C_4=2*Mo                               //[H_3O+](M)
 pH_4=-log10(C_4)
 disp(pH_4,'pH of the given solution')
diff --git a/3428/CH23/EX14.23.18/Ex14_23_18.sce b/3428/CH23/EX14.23.18/Ex14_23_18.sce
index a19fbf110..0fdda2abd 100644
--- a/3428/CH23/EX14.23.18/Ex14_23_18.sce
+++ b/3428/CH23/EX14.23.18/Ex14_23_18.sce
@@ -1,21 +1,21 @@
 //Section-14,Example-2,Page no.-PC.112
 //To calculate the pH in the following cases.
 clc;
-V_1=150                //volume of 0.1 NaOH solution
-V_2=150                //volume of 0.2 HCl solution
-N_1=0.1
-N_2=0.2
-V=V_1+V_2             //Total volume of the solution
-m_eq=(V_2*N_2)-(V_1*N_1)            //Total milliequivalents of excess HCl
-N=m_eq/V
-C_1=N                          //Since HCl is a strong acid so[HCl]=[H3O+]
+V_1=150                //volume of 0.1 NaOH solution(ml)
+V_2=150                //volume of 0.2 HCl solution(ml)
+N_1=0.1                 //(N)
+N_2=0.2                    //(N)
+V=V_1+V_2             //Total volume of the solution(ml)
+m_eq=(V_2*N_2)-(V_1*N_1)            //Total milligram equivalents of excess HCl(gm equivalents)
+N=m_eq/V                              //(N)
+C_1=N                          //Since HCl is a strong acid so[HCl]=[H3O+]   (M)
 pH_1=-log10(C_1)
 disp(pH_1,'pH of the required solution')
 pH1=5
-C1=10^-5                    //[H3O+]
+C1=10^-5                    //[H3O+]             (M)
 pH2=3
-C2=10^-3                    //[H3O+]
-C_3=(C1+C2)/2                //[H3O+]
+C2=10^-3                    //[H3O+]                (M)
+C_3=(C1+C2)/2                //[H3O+]                  (M)
 pH_2=-log10(C_3)
 disp(pH_2,'pH of the required solution')
 
diff --git a/3428/CH23/EX14.23.2/Ex14_23_2.sce b/3428/CH23/EX14.23.2/Ex14_23_2.sce
index f4f950dd0..5bd1ce993 100644
--- a/3428/CH23/EX14.23.2/Ex14_23_2.sce
+++ b/3428/CH23/EX14.23.2/Ex14_23_2.sce
@@ -1,9 +1,9 @@
 //Section-14,Example-1,Page no.-PC.71

 //To find Q_c and predict the direction in which the reaction would proceed.

 clc;

-[N_2]=0.03

-[O_2]=0.01

-[NO]=0.04

+[N_2]=0.03                   //mol

+[O_2]=0.01                    //mol

+[NO]=0.04                     //mol

 K_c=2.2*10^3

 Q_c=([N_2]*[O_2])/[NO]^2

 disp(Q_c)

diff --git a/3428/CH23/EX14.23.20/Ex14_23_20.sce b/3428/CH23/EX14.23.20/Ex14_23_20.sce
index 4535b7238..bf7f8f1cd 100644
--- a/3428/CH23/EX14.23.20/Ex14_23_20.sce
+++ b/3428/CH23/EX14.23.20/Ex14_23_20.sce
@@ -2,15 +2,15 @@
 //To calculate the pH in the following cases.
 clc;
 K_a=7.3*10^-6
-c_1=0.23
+c_1=0.23                 //(M)
 alpha_1=sqrt(K_a/c_1)
-C_1=c_1*alpha_1
+C_1=c_1*alpha_1           //(M)
 pH_1=-log10(C_1)
 disp(pH_1,'pH of the given weak acid')
-c_2=0.2
+c_2=0.2                   //(M)
 K_b=4.4*10^-5
 alpha_2=sqrt(K_b/c_2)
-C_2=c_2*alpha_2                 //[OH-]
+C_2=c_2*alpha_2                 //[OH-] (M)
 pOH=-log10(C_2)
 pH_2=14-pOH
 disp(pH_2,'pH of CH_3NH_2')
diff --git a/3428/CH23/EX14.23.21/Ex14_23_21.sce b/3428/CH23/EX14.23.21/Ex14_23_21.sce
index 19726bf37..34102a8e6 100644
--- a/3428/CH23/EX14.23.21/Ex14_23_21.sce
+++ b/3428/CH23/EX14.23.21/Ex14_23_21.sce
@@ -1,12 +1,13 @@
 //Section-14,Example-5,Page no.-PC.114
 //To calculate the pH of a solution obtained in the given condition.
 clc;
-M_1=0.1
-M_2=0.2
-V_1=10
-V_2=40
-V=V_1+V_2
+M_1=0.1                                 //(M)
+M_2=0.2                                     //(M)
+V_1=10                            //(ml)
+V_2=40                              //(ml)
+V=V_1+V_2                      //(ml)
 m_eq=(M_1*V_1)+(M_2*V_2*2)             //mgm. equivalents of [H3O+]
-C_1=m_eq/V                   //[H3O+]
+C_1=m_eq/V                   //[H3O+] //(M)
 pH=-log10(C_1)
 disp(pH,'pH of the given solution')
+//Answer given in the book(pH=4.685) is wrong
diff --git a/3428/CH23/EX14.23.22/Ex14_23_22.sce b/3428/CH23/EX14.23.22/Ex14_23_22.sce
index 32f3b36a6..c3db5aa20 100644
--- a/3428/CH23/EX14.23.22/Ex14_23_22.sce
+++ b/3428/CH23/EX14.23.22/Ex14_23_22.sce
@@ -1,8 +1,11 @@
 //Section-14,Example-6,Page no.-PC.114
 //To calculate the pH of 10^-8 M HCl solution.
 clc;
+C=10^-8    //(M) Concentration of HCl solution 
 k_w=10^-14
-x=9.5*10^-8
-C_1=10^-8+x
+x1=(-C+sqrt((C)^2-(4*1*(-k_w))))/2                       //(M) Concentration of OH-
+x2=(-C-sqrt((C)^2+(4*1*(-k_w))))/2                       //(M) Concentration of OH- 
+//since value of x2 is complex so it is rejected.
+C_1=C+x1                         //(M) Concentration of (H3O+)
 pH=-log10(C_1)
 disp(pH,'pH of the given solution')
diff --git a/3428/CH23/EX14.23.23/Ex14_23_23.sce b/3428/CH23/EX14.23.23/Ex14_23_23.sce
index ea1a8508b..53224b6ae 100644
--- a/3428/CH23/EX14.23.23/Ex14_23_23.sce
+++ b/3428/CH23/EX14.23.23/Ex14_23_23.sce
@@ -5,5 +5,5 @@ pKa=3.08
 Ka=10^(-pKa)
 x1=(-Ka+sqrt((Ka)^2+(4*0.01*Ka)))/2
 x2=(-Ka-sqrt((Ka)^2+(4*0.01*Ka)))/2          //neglected
-pH=-log10(x2)
+pH=-log10(x1)
 disp(pH,'pH of the given solution')
diff --git a/3428/CH23/EX14.23.25/Ex14_23_25.sce b/3428/CH23/EX14.23.25/Ex14_23_25.sce
index d66e1afac..3f7a15338 100644
--- a/3428/CH23/EX14.23.25/Ex14_23_25.sce
+++ b/3428/CH23/EX14.23.25/Ex14_23_25.sce
@@ -2,7 +2,7 @@
 //To estimate the pH of 0.01 M CH3COONa.

 clc;

 pK_w=14

-B=0.01

+B=0.01                //(M)

 pK_a=-log10(1.8*10^-5)

 pH=(1/2*pK_w)+(1/2*log10(B))+(1/2*pK_a)

 disp(pH,'pH of given solution of CH3COONa')

diff --git a/3428/CH23/EX14.23.26/Ex14_23_26.sce b/3428/CH23/EX14.23.26/Ex14_23_26.sce
index bd8cf4310..443e96eb1 100644
--- a/3428/CH23/EX14.23.26/Ex14_23_26.sce
+++ b/3428/CH23/EX14.23.26/Ex14_23_26.sce
@@ -3,7 +3,7 @@
 clc;

 K_a=1.75*10^-5

 pK_a=-log10(K_a)

-[CH_3COOH]=(1000/(60*100))

-[CH_3COONa]=((1.5*1000)/(82*100))

+[CH_3COOH]=(1000/(60*100))               //moldm^-3

+[CH_3COONa]=((1.5*1000)/(82*100))           //moldm^-3

 pH=(pK_a+ (log10([CH_3COONa]/[CH_3COOH])))

 disp(pH,'pH of the given solution')

diff --git a/3428/CH23/EX14.23.27/Ex14_23_27.sce b/3428/CH23/EX14.23.27/Ex14_23_27.sce
index c494a9345..8728c5de9 100644
--- a/3428/CH23/EX14.23.27/Ex14_23_27.sce
+++ b/3428/CH23/EX14.23.27/Ex14_23_27.sce
@@ -1,19 +1,19 @@
 //Section-14,Example-2,Page no.-PC.126

 clc;

-CH_3COONa_1=0.01

-CH_3COOH_1=0.1

+CH_3COONa_1=0.01                 //moldm^-3

+CH_3COOH_1=0.1                   //moldm^-3

 K_a=1.75*10^-5

 pK_a=-log10(K_a)

 pH1=pK_a +(log10(CH_3COONa_1/CH_3COOH_1))

 disp(pH1,'pH of the given buffer solution')

-HCl=0.0002

-CH_3COONa_2=0.01+0.0002

-CH_3COOH_2=0.1-0.002

+HCl=0.0002                        //moles

+CH_3COONa_2=0.01+0.0002            // moldm^-3

+CH_3COOH_2=0.1-0.002               //moldm^-3

 pH2=pK_a +(log10(CH_3COONa_2/CH_3COOH_2))

 disp(pH2,'pH of the solution after addition of HCl')

-C_1=pH1-pH2                   //change in pH

-CH_3COONa_3=0.01+0.002

-CH_3COOH_3=0.1-0.002

+C_1=pH1-pH2                   //change in pH (M)

+CH_3COONa_3=0.01+0.002                //moldm^-3

+CH_3COOH_3=0.1-0.002                  //moldm^-3

 pH3=pK_a +(log10(CH_3COONa_3/CH_3COOH_3))

 pH4=pH1-pH3                //change in pH

 disp(pH4,'Required pH')

diff --git a/3428/CH23/EX14.23.30/Ex14_23_30.sce b/3428/CH23/EX14.23.30/Ex14_23_30.sce
index 157ef4c65..47a2f1098 100644
--- a/3428/CH23/EX14.23.30/Ex14_23_30.sce
+++ b/3428/CH23/EX14.23.30/Ex14_23_30.sce
@@ -1,10 +1,10 @@
 //Section-14,Example-2,Page no.-PC.129

 //To find the Molarity of given sulphuric acid solution.

 clc;

-N_T=0.1354

-V_T=42.20

-V_S=50.00

+N_T=0.1354             //(N)

+V_T=42.20             //(ml)

+V_S=50.00              //(ml)

 //N_S=N_H_2SO_4(let) and M_S=M_H_2SO_4

-N_S=(N_T*V_T)/V_S

-M_S=(N_S/2)

+N_S=(N_T*V_T)/V_S         //(N)

+M_S=(N_S/2)               //(M)

 disp(M_S,'Molarity of H_2SO_4')

diff --git a/3428/CH23/EX14.23.4/Ex14_23_4.sce b/3428/CH23/EX14.23.4/Ex14_23_4.sce
index 75861f586..798ba3942 100644
--- a/3428/CH23/EX14.23.4/Ex14_23_4.sce
+++ b/3428/CH23/EX14.23.4/Ex14_23_4.sce
@@ -1,12 +1,19 @@
 //Section-14,Example-2,Page no.-PC.75

 //To calculate equilibrium composition of reaction mixture

 clc;

-a_N2=1.00-x

-a_H2=3.00-(3*x)

-a_NH_3=2*x

+//a_N2=1.00-x           //Equilibrium partial pressure of N_2(bar)

+//a_H2=3.00-(3*x)          //Equilibrium partial pressure of H_2(bar)

+//a_NH_3=2*x               //Equilibrium partial pressure of NH(bar)

 //K=(2*x^2)/((1.00-x)*(3.00-3*x)^3)=977

 x_1=(163.416+sqrt((163.416)^2-(4*81.208*81.208)))/(2*81.208)

 x_2=(163.416-sqrt((163.416)^2-(4*81.208*81.208)))/(2*81.208)

-p_N2=1-x_2

-p_H2=3*(1-x_2)

-p_NH3=2*x_2

+disp(x_1,'1st value of K')

+disp(x_2,'2nd value of K')

+//since x_2<1,it is accepted as the value of x.

+p_N2=1-x_2              //bar

+disp(p_N2,'Equilibrium composition of N2(bar)')

+p_H2=3*(1-x_2)           //bar

+disp(p_H2,'Equilibrium composition of H2(bar)')

+p_NH3=2*x_2                 //bar

+disp(p_NH3,'Equilibrium composition of NH3(bar)')

+// It can be concluded that product NH3 dominate at equilibrium.

diff --git a/3428/CH23/EX14.23.7/Ex14_23_7.sce b/3428/CH23/EX14.23.7/Ex14_23_7.sce
index 87f2a4e84..a2eed29b4 100644
--- a/3428/CH23/EX14.23.7/Ex14_23_7.sce
+++ b/3428/CH23/EX14.23.7/Ex14_23_7.sce
@@ -1,11 +1,11 @@
 //Section-14,Example-1,Page no.-PC.80

 //To calculate K_p at 1000 K

 clc;

-T_1=925

-T_2=1000

-K_p925=18.5

-dl_H=-71.09*10^3

-R=8.314

+T_1=925                          //Kelvin

+T_2=1000                         //Kelvin

+K_p925=18.5                        //K_p at 925 K

+dl_H=-71.09*10^3                  //kJmol^-1

+R=8.314                            //JK^-1mol^-1

 //ln(K_p1000)/(K_p925)=(dl_H/R)*((1/T_1)-(1/T_2))

 K=((dl_H)/R)*((1/T_1)-(1/T_2))

 K_p1000=(%e^(K))*18.5

diff --git a/3428/CH23/EX14.23.9/Ex14_23_9.sce b/3428/CH23/EX14.23.9/Ex14_23_9.sce
index 9d2925809..927bbbd96 100644
--- a/3428/CH23/EX14.23.9/Ex14_23_9.sce
+++ b/3428/CH23/EX14.23.9/Ex14_23_9.sce
@@ -1,6 +1,6 @@
 //Section-14,Example-4,Page no.-PC.82

 //To calculate the solubility of Ag_2CrO_4.

 clc;

-K_sp=(9*10^-12)/4

-S=(K_sp)^(1/3)

+K_sp=(9*10^-12)

+S=(K_sp/4)^(1/3)

 disp(S,'Solubility product of Ag_2CrO_4(mol/dm^3)')