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-//Chapter-8,Example 3,Page 195
-clc();
-close();
-
-//solution for (a) part
-
-conc1=1*10^-8 //concentration of HCl solution
-
-//let [H+] concentration from water = x
-//so, [H+] of solution = conc1*x an [OH-] = x
-//......Kw = [H+]*[OH-] = 10^-14
-//......x^2 +(10^-8)*x -(10^-14)=0
-x = (-10^-8 + sqrt((10^-8)^2 + 4*1*10^-14))/(2*1)
-
-H=conc1 +x
-
-pH1=-log10(H)
-
-printf('for HCl the pH = %.3f',pH1)
-
-
-//solution for (b) part
-conc2= 1*10^-8 //concentration of NaOH solution
-
-OH=x+conc2
-
-pOH2=-log10(OH)
-
-pH2=14 - pOH2
-
-printf('\n for NaOH the pH = %.3f',pH2)
-