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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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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)