## Chemistry (7th Edition)

a) $[OH^-] = 4 \times 10^{- 11}M$, Acidic. b)$[OH^-] = 5 \times 10^{- 15}M$, Acidic. c) $[H_3O^+] = 1.786 \times 10^{- 6}M$, Acidic. d)$[H_3O^+] = 6.667 \times 10^{- 12}M$, Basic. e) $[H_3O^+] = 1 \times 10^{- 7}M$, Neutral.
If $[H_3O^+] > [OH^-]$, the solution is acidic. If $[H_3O^+] < [OH^-]$, the solution is basic. If $[H_3O^+] = [OH^-]$, the solution is neutral. a) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $2.5 \times 10^{- 4} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 2.5 \times 10^{- 4}}$ $[OH^-] = 4 \times 10^{- 11}$ - Acidic b) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $2.0 * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 2.0}$ $[OH^-] = 5 \times 10^{- 15}$ - Acidic c) $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $5.6 \times 10^{- 9} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 5.6 \times 10^{- 9}}$ $[H_3O^+] = 1.786 \times 10^{- 6}$ - Acidic d) $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $1.5 \times 10^{- 3} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 1.5 \times 10^{- 3}}$ $[H_3O^+] = 6.667 \times 10^{- 12}$ - Basic e) $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $1 \times 10^{- 7} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 1 \times 10^{- 7}}$ $[H_3O^+] = 1 \times 10^{- 7}$ - Neutral