## Chemistry (7th Edition)

a)$[OH^-] = 2.941 \times 10^{- 6}M$ b)$[H_3O^+] = 1 \times 10^{- 12}M$ c)$[H_3O^+] = 1 \times 10^{- 4}M$ d)$[OH^-] = 1 \times 10^{- 7}M$ e)$[OH^-] = 1.163 \times 10^{- 10}M$
- 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) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $3.4 \times 10^{- 9} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 3.4 \times 10^{- 9}}$ $[OH^-] = 2.941 \times 10^{- 6}$ - Basic b) $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $1 \times 10^{- 2} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 1 \times 10^{- 2}}$ $[H_3O^+] = 1 \times 10^{- 12}$ - Basic c) $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $1 \times 10^{- 10} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 1 \times 10^{- 10}}$ $[H_3O^+] = 1 \times 10^{- 4}$ - Acidic d) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $1 \times 10^{- 7} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 1 \times 10^{- 7}}$ $[OH^-] = 1 \times 10^{- 7}$ - Neutral e) $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $8.6 \times 10^{- 5} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 8.6 \times 10^{- 5}}$ $[OH^-] = 1.163 \times 10^{- 10}$ - Acidic