if you have a solution that mixes 25.0 mL of 0.200 M HNO2 with 75.0 mL of 0.160 M RbNO2, what is the resulting pH?

Question

If you have a solution that mixes 25.0 mL of 0.200 M HNO₂ with 75.0 mL of 0.160 M RbNO2, what is the resulting pH? (KNO₂ Ka = 4.5 x 10^-4)

Answer 1

The concept here is that the resulting solution contains a weak acid (HNO₂) and its conjugate base (NO₂^-, the Rb⁺ is a neutral counterion and is not important to the question).

This type of solution is often called a buffer. The pH will be close to the pKa, and the exact pH can be calculated using the Henderson-Hasselbalch equation

pH = pKa + log(B/A) (A = acid, B = base). You can use concentrations or number of mol in the equation for A and B.

Mol of acid: 0.025 L x (0.200 mol/L) = 0.005 mol

Mol of base: 0.075 L x (0.160 mol/L) = 0.012 mol

Note that there is more base than acid, so we expect the pH to be a little higher than pKa

pKa = -log(4.5 x 10^-4) = 3.35

pH = 3.35 + log(0.012/0.005) = 3.35 + 0.38 = 3.73