JAMB - Chemistry (2008 - No. 18)
What is the pH of 0.001 mol dm-3 solution of sodium hydroxide
14
13
12
11
Explication
\([H^+] [OH^-] = 10^{-14}\)
\([H^+] = 10^{-14}/[OH^-] = 10^{-14}/(1 * 10^{-3}) = 10^{-11}\)
pH = -Log10[H+]
= - Log10[10-11] = - 1 * -11 Log10[10-11] = 1 = -1 * -11 * 1 = 11
\([H^+] = 10^{-14}/[OH^-] = 10^{-14}/(1 * 10^{-3}) = 10^{-11}\)
pH = -Log10[H+]
= - Log10[10-11] = - 1 * -11 Log10[10-11] = 1 = -1 * -11 * 1 = 11