## Chemistry: Molecular Approach (4th Edition)

a. $[H_3O^+] = 9.1 \times 10^{-6}$, the solution is Acidic. b. $[H_3O^+] = 3.4 \times 10^{-13}$, the solution is Basic. c. $[H_3O^+] = 1.4 \times 10^{-3}$, the solution is Acidic.
1. The $K_w$ at $25 ^oC$ is equal to $10^{-14}$. Using that, calculate $[H_3O^+]$ a. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.1 \times 10^{-9} } = 9.1 \times 10^{-6} \space M$$ b. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 2.9 \times 10^{-2} } = 3.4 \times 10^{-13} \space M$$ c.$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 6.9 \times 10^{-12} } = 1.4 \times 10^{-3} \space M$$ 2. If $[H_3O^+] \gt [OH^-]$, the solution is acidic; if $[H_3O^+] \lt [OH^-]$, the solution is basic. a. $[H_3O^+] \gt [OH^-]$: Acidic. b. $[H_3O^+] \lt [OH^-]$: Basic. c. $[H_3O^+] \gt [OH^-]$: Acidic.