ph of resultant solution formed by mixing of equal volumes of two solutions having ph 5.8 and ph 12.8

ph of resultant solution formed by mixing of equal volumes of two solu

1.	ph of resultant solution formed by mixing of equal volumes of two solutions having ph 5.8 and ph 12.8
Answer» First of all, we have to find out CONCENTRATION of hydrogen ions in both solutions. we know from Arrhenius theory, $pH=-<klux>LOG</klux>[H^+]$ 5.8 = -log[H^+] log[H^+] = -5.8 [H^+] = antilog( -5.8) = 1.58489 × 10^-6 SIMILARLY, pH = 12.8 -log[H^+] = 12.8 [H^+] = antilog(-12.8) = 1.58489 × 10^-13 now, use formula, $M_1V_1+M_2V_2=M(V_1+V_2)$ here, V1 and V2 are volumes of each solutions a/c to question, V1 = V2 = V (let) then, 1.58489 × 10^-6 × V + 1.58489 × 10^-13 × V = M × 2V (1.58489 × 10^-6 + 1.58489 × 10^-13)/2 = M M = 7.92445 × 10^-7 now, pH = -log(7.92445 × 10^-7) = 6.1