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-//Chapter-8,Example 4,Page 196
-clc();
-close();
-
-alpha1=0.02
-
-Ka=1.8*10^-5
-
-//at equilibrium..
-//[CH3COOH] = C1* (1-alpha1)
-//[H+] = C1* alpha1
-//[CH3COO-] = C1* alpha1
-// Ka =[H+] * [CH3COO-]/[CH3COOH]
-// Ka = C1* alpha1*C1* alpha1/(C1 (1-alpha1))
-
-C1=Ka*(1-alpha1)/alpha1^2
-
-printf('the molar concentration of CH3COOH is C = %.4f molar',C1)
-
-C2=0.01
-
-alpha2= sqrt(Ka/C2)
-
-printf('\n alpha = %.4f ',alpha2)