Answer
\begin{vmatrix}
[H_3O^+] & [OH^-] & pH & Acidic \space or Basic \\
3.5 \times 10^{-3} & 2.9 \times 10^{-12} & 2.46 & Acidic \\
2.6 \times 10^{-8} & 3.8 \times 10^{-7} & 7.59 & Basic\\
1.8 \times 10^{-9} & 5.6 \times 10^{-6} & 8.74 & Basic \\
7.1 \times 10^{-8} & 1.4 \times 10^{-7} & 7.15 & Basic
\end{vmatrix}
Work Step by Step
We find:
a. $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 3.5 \times 10^{-3} } = 2.9 \times 10^{-12} \space M$$
$$pH = -log[H_3O^+] = -log( 3.5 \times 10^{-3} ) = 2.46 $$
pH < 7: Acidic
b. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 3.8 \times 10^{-7} } = 2.6 \times 10^{-8} \space M$$
$$pH = -log[H_3O^+] = -log( 2.6 \times 10^{-8} ) = 7.59 $$
pH > 7: Basic
c. $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 1.8 \times 10^{-9} } = 5.6 \times 10^{-6} \space M$$
$$pH = -log[H_3O^+] = -log( 1.8 \times 10^{-9} ) = 8.74 $$
pH > 7: Basic
d. $$[H_3O^+] = 10^{-pH} = 10^{-7.15} = 7.1 \times 10^{-8} \space M$$
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 7.1 \times 10^{-8} } = 1.4 \times 10^{-7} \space M$$
pH > 7: Basic
