## Chemistry 10th Edition

$[H_3O^+] = 0.0548M$ $[OH^-] = 1.825 \times 10^{-13}M$ $pH = 1.261$ $pOH = 12.739$
1. HCl is a strong acid, so : $[HCl] = [H_3O^+] = 0.0548M:$ 2. Now, find the pH and pOH: $pH = -log[H_3O^+]$ $pH = -log( 0.0548)$ $pH = 1.261$ $pOH + pH = 14$ $pOH = 14 - pH = 14 - 1.261$ $pOH = 12.739$ 3. And the $[OH^-]$: $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $5.48 \times 10^{- 2} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 5.48 \times 10^{- 2}}$ $[OH^-] = 1.825 \times 10^{- 13}M$