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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)
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