Calculate the pH during the titration of 20.0 mL of 0.30 M HNO3

Question

Calculate the pH during the titration of 20.0 mL of 0.30 M HNO₃(aq) with 0.30 M KOH after 19.3 mL of the base have been added.

My answer was 11.73

The correct answer is 2.27

Can you explain why the answer is 2.27 and not 11.73

Answer 1

Hi Oscar,

Great question. Are you familiar with constructing ICE tables? This will be the necessary strategy to solve this problem.

If not, the ICE stands for initial, change, and equilibrium. These acid and base problems are all about how the system reaches equilibrium after neutralization occurs (H + OH = H₂O)

Let's track what's happening to the HNO₃ as soon as we add the KOH. We will construct the ICE table using moles because this is often the easier unit to work with.

Equation:

HNO₃ + KOH -> H₂O + KNO₃

Initial:

mol of HNO₃ = 0.30 M * 0.020 L = 0.006 mol

mol of KOH = 0.30 M * 0.0193 L = 0.00579 mol

We do not need to account for change in H₂O since it is a liquid

mol of KNO₃ = 0 mol

Change:

Our limiting reagent (the one consumed first) is KOH, so 0.00579 mol KOH will neutralize 0.00579 mol of HNO₃. There will be 0.00579 mol of KNO₃ generated.

Equilibrium:

mol HNO₃ = 0.00021 mol

mol KOH = 0 mol (consumed)

mol KNO₃ = 0.00579 mol

Since KNO₃ is not an acid or base, it does not contribute to the overall pH. Therefore, we will calculate the pH by the concentration of H⁺ we have in solution. We used up all of the OH^- to neutralize the HNO₃.

mol HNO₃ = mol H⁺

pH = -log[H⁺]

[H⁺] = mol HNO₃/L solution = 0.00021 mol/ (0.020 L + 0.0193 L) = 0.00534 M

pH = -log[0.00534 M] = 2.27

Remember that the concentration will be dictated by the final volume of the solution.

*It looks like you just flipped which ion was left in solution after the titration. I think you found that the [KOH] was 0.00534 M versus the [HNO₃], which gives a pH of 11.73. Be sure to keep track of which species is the limiting reagent in the neutralization.

Any more questions? Let me know!

Thank you,

Jacob P.