Answer
\begin{matrix}
Food & [H_3O^+] & [OH^-] & pH & Acidic, Basic \space or \space Neutral? \\
Strawberries & 1.3 \times 10^{-4} \space M & 7.7 \times 10^{-11} \space M & 3.90 & Acidic \\
Soy \space milk & 1.0 \times 10^{-7} \space M & 1.0 \times 10^{-7} \space M & 7.00 & Neutral \\
Tuna \space fish & 6.3 \times 10^{-7} \space M & 1.6 \times 10^{-8} \space M & 6.20 & Acidic
\end{matrix}
Work Step by Step
Strawberries:
$$[H_3O^+] = 10^{-pH} = 10^{-3.90} = 1.3 \times 10^{-4} \space M$$
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 1.3 \times 10^{-4} } = 7.7 \times 10^{-11} \space M$$
Since $[H_3O^+] \gt [OH^-]$ and $pH \lt 7$, this solution is acidic.
Soy milk:
All neutral solutions have:
$pH = 7.00$
$[H_3O^+] = [OH^-] = 1.0 \times 10^{-7} \space M$
Tuna fish:
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 6.3 \times 10^{-7} } = 1.6 \times 10^{-8} \space M$$ $$pH = -log[H_3O^+] = -log( 6.3 \times 10^{-7} ) = 6.20 $$
Since $[H_3O^+] \gt [OH^-]$ and $pH \lt 7$, this solution is acidic.