Answer
\begin{matrix}
Food & [H_3O^+] & [OH^-] & pH & Acidic, Basic \space or \space Neutral? \\
Rye \space bread & 1.6 \times 10^{-9} \space M & 6.3 \times 10^{-6} \space M & 8.8 & Basic \\
Tomatoes & 2.3 \times 10^{-5} \space M & 4.3 \times 10^{-10} \space M & 4.64 & Acidic \\
Peas & 6.2 \times 10^{-7} \space M & 1.6 \times 10^{-8} \space M & 6.21 & Acidic
\end{matrix}
Work Step by Step
Rye bread:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 6.3 \times 10^{-6} } = 1.6 \times 10^{-9} \space M$$
$$pH = -log[H_3O^+] = -log( 1.6 \times 10^{-9} ) = 8.8 $$
Since $[OH^-] \gt [H_3O^+]$ and $pH \gt 7$, this solution is basic.
Tomatoes:
$$[H_3O^+] = 10^{-pH} = 10^{-4.64} = 2.3 \times 10^{-5} \space M$$
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 2.3 \times 10^{-5} } = 4.3 \times 10^{-10} \space M$$
Since $[H_3O^+] \gt [OH^-]$ and $pH \lt 7$, this solution is acidic.
Peas:
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 6.2 \times 10^{-7} } = 1.6 \times 10^{-8} \space M$$
$$pH = -log[H_3O^+] = -log( 6.2 \times 10^{-7} ) = 6.21 $$
Since $[H_3O^+] \gt [OH^-]$ and $pH \lt 7$, this solution is acidic.