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diff --git a/2465/CH8/EX8.4/Example_4.sce b/2465/CH8/EX8.4/Example_4.sce
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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)