How many moles of H3O^+ or OH^- must you add to 5.6 L of HA solution to adjust its pH from 4.52 to 5.25? Assume a negligible volume change.

Here's how to solve these problems:

18.20 How many moles of $H_3O^+$ or $OH^-$ must you add to 5.6 L of HA solution to adjust its pH from 4.52 to 5.25? Assume a negligible volume change.

Step 1: Determine the initial and final hydronium ion concentrations.
The initial pH is 4.52, so the initial hydronium ion concentration is:

$$[H_3O^+]_1 = 10^{-pH_1} = 10^{-4.52} \text{ M} = 3.01995 \times 10^{-5} \text{ M}$$

The final pH is 5.25, so the final hydronium ion concentration is:

$$$$