## Chemistry: Molecular Approach (4th Edition)

a. $[H_3O^+] = 6.7 \times 10^{-3}$: Basic solution. a. $[H_3O^+] = 1.0 \times 10^{-7}$: Neutral solution. a. $[H_3O^+] = 1.2 \times 10^{-5}$: Acidic solution.
a. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.5 \times 10^{-2} } = 6.7 \times 10^{-13} \space M$$ Since $[OH^-] \gt [H_3O]$, this solution is basic. b. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.0 \times 10^{-7} } = 1.0 \times 10^{-7} \space M$$ Since $[OH^-]=[H_3O^+]$, this solution is neutral. c.$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 8.2 \times 10^{-10} } = 1.2 \times 10^{-5} \space M$$ Since $[H_3O^+]\gt[OH^-]$, this solution is acidic.