## Chemistry: Atoms First (2nd Edition)

a. $[H^+]$ and $[OH^-] = 1.709\times 10^{- 7}$ b. It is equal to $6.767$. c. $pH$ = $12.530$
a. - Pure water solution : $[H^+] = [OH^-]$ - Create a unknown called 'x', that has the value of $[H^+]\ and\ [OH^-]$ - Write the $K_w$ expression, and calculate the x value: $[H^+] * [OH^-] = K_w = 2.92\times 10^{- 14}$ $x * x = 2.92\times 10^{- 14}$ $x^2 = 2.92\times 10^{- 14}$ $x = \sqrt { 2.92\times 10^{- 14}}$ $x = 1.71\times 10^{- 7}$ Therefore: $[H^+]\ and\ [OH^-] = 1.71\times 10^{- 7}M$ b. $pH = -log[H^+]$ $pH = -log( 1.71 \times 10^{- 7})$ $pH = 6.767$ c. $[OH^-] * [H^+] = Kw = 2.92 \times 10^{-14}$ $0.1 * [H^+] = 2.92 \times 10^{-14}$ $[H^+] = \frac{2.92 \times 10^{-14}}{0.1}$ $[H^+] = 2.92 \times 10^{-13}M$ $pH = -log[H^+]$ $pH = -log( 2.92 \times 10^{- 13})$ $pH = 12.53$