10 files changed, 142 insertions, 0 deletions
diff --git a/3856/CH11/EX11.1/Ex11_1.sce b/3856/CH11/EX11.1/Ex11_1.sce
new file mode 100644
index 000000000..d71045216
--- /dev/null
+++ b/3856/CH11/EX11.1/Ex11_1.sce
@@ -0,0 +1,23 @@
+//Calculate the concentration of the undissociated acid ,the H positive ion and the CN negative ion .And the percent dissociation
+
+//Example 11.1
+
+clc;
+
+clear;
+
+Ka=4.9*10^-10;  //Dissociatin constant of weak acid HCN at 298 K 
+
+x1=0.050;  //Concentration of HCN in M ,(HCN is a aweak acid assuming that at equilibrium the undissociated molecule of HCN is also same )
+
+x=(Ka*x1)^(1/2);  //Concentration of H ion and CN ion at equilibrium in M (cocentration of both ion is equal)
+
+printf("Concentration of ion = %.0f*10^-6 M",x*10^6);
+
+x2=x1-x; //Concentration of undissociated acid at equilibrium in M
+
+printf("\n Concentration of undissociated acid at equilibrium = %.3f M",x2)
+
+X=(x/x1)*100;  //Percent dissociation of HCN 
+
+printf("\nPercent dissociation = %.0f*10^-2 percent ",X*10^2);
diff --git a/3856/CH11/EX11.1/Ex11_1.txt b/3856/CH11/EX11.1/Ex11_1.txt
new file mode 100644
index 000000000..b95ee2acf
--- /dev/null
+++ b/3856/CH11/EX11.1/Ex11_1.txt
@@ -0,0 +1,3 @@
+ Concentration of ion = 5*10^-6 M
+ Concentration of undissociated acid at equilibrium = 0.050 M
+Percent dissociation = 1*10^-2 percant 
+\ No newline at end of file
diff --git a/3856/CH11/EX11.2/Ex11_2.sce b/3856/CH11/EX11.2/Ex11_2.sce
new file mode 100644
index 000000000..006c85cfe
--- /dev/null
+++ b/3856/CH11/EX11.2/Ex11_2.sce
@@ -0,0 +1,19 @@
+//Monitoring the Titration
+
+//Example 11.2
+
+clc;
+
+clear;
+
+Kin=4*10^-10;//Equilibrium Constant
+
+pKin=-log10(Kin);//Negative Logarithm of Kin
+
+phl=pKin-1;//Lower Value of pH
+
+phu=pKin+1;//Upper Value of pH
+
+printf("Phenophthalein can be used as an indicator as it begins to change color from acid(colourless) at pH %f",phl);
+
+printf("\nto base form (reddish pink)at pH %f",phu)
diff --git a/3856/CH11/EX11.2/Ex11_2.txt b/3856/CH11/EX11.2/Ex11_2.txt
new file mode 100644
index 000000000..eef5c5fde
--- /dev/null
+++ b/3856/CH11/EX11.2/Ex11_2.txt
@@ -0,0 +1,2 @@
+ Phenophthalein can be used as an indicator as it begins to change color from acid(colourless) at pH 8.397940
+to base form (reddish pink)at pH 10.397940
+\ No newline at end of file
diff --git a/3856/CH11/EX11.3/Ex11_3.sce b/3856/CH11/EX11.3/Ex11_3.sce
new file mode 100644
index 000000000..a9ada49ac
--- /dev/null
+++ b/3856/CH11/EX11.3/Ex11_3.sce
@@ -0,0 +1,37 @@
+//To Find the Concentrations of all the species in the reaction
+
+//Example 11.3
+
+clc;
+
+clear;
+
+Ka=4.2*10^-7;//Acid Dissociation Constant for Carbonic Acid
+
+Sol=1.1*10^-5;//Solubility of CO2 in equilibrium with water
+
+a1=1;b1=Ka;c1=-Ka*Sol;//Coefficients a,b and c of the quadratic equation to find the concentration of H+
+
+d1=(b1^2-(4*a1*c1));//Discriminant of the Quadratic Equation
+
+x=(-b1+sqrt(d1))/(2*a1);//Concentration of H+
+
+Ka2=4.8*10^-11;//Second Dissociation Constant for H2CO3
+
+y=Ka2;//Concentration of CO3 2- ions
+
+Kw=1*10^-14;//Disscociation Constant of Water
+
+z=Kw/x;//Concentration of OH- ions (The answer vary due to round off error)
+
+printf("At Equilibrium the concentrations are as follows:");
+
+printf("\n [H+]=%.1f*10^-6 M",x*10^6);
+
+printf("\n [OH-]=%.1f*10^-9 M",z*10^9);
+
+printf("\n [H2CO3]=%.1f*10^-5 M",Sol*10^5);
+
+printf("\n [HCO3-]=%.1f*10^-6 M",x*10^6);
+
+printf("\n [CO3 2-]=%.1f*10^-11 M",y*10^11);
diff --git a/3856/CH11/EX11.3/Ex11_3.txt b/3856/CH11/EX11.3/Ex11_3.txt
new file mode 100644
index 000000000..885d843fd
--- /dev/null
+++ b/3856/CH11/EX11.3/Ex11_3.txt
@@ -0,0 +1,6 @@
+ At Equilibrium the concentrations are as follows:
+ [H+]=1.9*10^-6 M
+ [OH-]=5.1*10^-9 M
+ [H2CO3]=1.1*10^-5 M
+ [HCO3-]=1.9*10^-6 M
+ [CO3 2-]=4.8*10^-11 M
+\ No newline at end of file
diff --git a/3856/CH11/EX11.4/Ex11_4.sce b/3856/CH11/EX11.4/Ex11_4.sce
new file mode 100644
index 000000000..2ac54d7b1
--- /dev/null
+++ b/3856/CH11/EX11.4/Ex11_4.sce
@@ -0,0 +1,21 @@
+//Calculate the pH of of a buffer solution what is pH of the buffer solution after the addition of HCl
+
+//Example 11.4
+
+clc;
+
+clear;
+
+C1=0.40; //Concentration of Acetic acid in M
+
+C2=0.55; //Concentration of Sodium Acetate in M
+
+pH1=4.76+log10(C2/C1); //pH of the Buffer solution before addition of HCl
+
+printf("pH of the Buffer solution = %.2f",pH1);
+
+C3=0.10; //Concentration of HCl in M
+
+pH=4.76+log10((C2-C3)/(C1+C3)); // pH of the Buffer solution after addition of HCl
+
+printf("\n pH of the Buffer solution after addition of HCl = %.2f",pH);
diff --git a/3856/CH11/EX11.4/Ex11_4.txt b/3856/CH11/EX11.4/Ex11_4.txt
new file mode 100644
index 000000000..6ea319742
--- /dev/null
+++ b/3856/CH11/EX11.4/Ex11_4.txt
@@ -0,0 +1,2 @@
+ pH of the Buffer solution = 4.90
+ pH of the Buffer solution after addition of HCl = 4.71
+\ No newline at end of file
diff --git a/3856/CH11/EX11.5/Ex11_5.sce b/3856/CH11/EX11.5/Ex11_5.sce
new file mode 100644
index 000000000..150d241e3
--- /dev/null
+++ b/3856/CH11/EX11.5/Ex11_5.sce
@@ -0,0 +1,28 @@
+//Describe how you would prepare a phosphate buffer with a pH of seven point four
+
+//Example 11.5 
+
+clc;
+
+clear;
+
+Ka1=7.5*10^-3; //Equilibrium consatnt for H3PO4= H+ +H2PO4-
+
+pKa1=-log10(Ka1); //minus logerithm of Ka1
+
+Ka2=6.2*10^-8; //Equilibrium consatnt for H2PO4-= H+ +HPO4--
+
+pKa2=-log10(Ka2); //minus logerithm of Ka2
+
+Ka3=4.8*10^-13; //Equilibrium consatnt for HPO4-- = H+ +PO3---
+
+pKa3=-log10(Ka3); //minus logerithm of Ka3
+
+pH=7.40; //pH of the required buffer solution
+
+C1=10^(pH-pKa2); //Concentratin of required solution to prepare buffer solution of  pH of 7.40
+
+C=C1/1.0; //Ratio  of the required solution to prepare buffer solution of  pH of 7.40 
+
+printf("Ratio of the required solution = %.2f The buffer is dissolve to disodium hydrogen phosphate and sodium dihydrogen phosphate in a mole ratio of 1.5:1.0 ",C);
+
diff --git a/3856/CH11/EX11.5/Ex11_5.txt b/3856/CH11/EX11.5/Ex11_5.txt
new file mode 100644
index 000000000..c26d903e9
--- /dev/null
+++ b/3856/CH11/EX11.5/Ex11_5.txt
@@ -0,0 +1 @@
+ Ratio of the required solution = 1.56 The buffer is dissolve to disodium hydrogen phosphate and sodium dihydrogen phosphate in a mole ratio of 1.5:1.0 
+\ No newline at end of file