Question 3 0.4 pts What is the KHP: NaOH mole ratio for the reaction between sodium hydroxide and potassium hydrogen phthalate, according to the balanced chemical equation? (fill in each blank with the appropriate integer value) Question 4 mol KHP : mol NaOH 1 pts A student removes 10.00 mL of 0.558M KHP solution and titrates it with a sodium hydroxide solution of unknown concentration. If 13.46mL of NaOH was used to reach the endpoint, what is the concentration of the NaOH solution in units of Molarity? (Be prepared to submit your complete set of calculations.) Type only the numerical portion of your answer in the box below. Your answer will be in units of Molarity (M). Report your answer to 3 significant figures.
Question 3 0.4 pts What is the KHP: NaOH mole ratio for the reaction between sodium hydroxide and potassium hydrogen phthalate, according to the balanced chemical equation? (fill in each blank with the appropriate integer value) Question 4 mol KHP : mol NaOH 1 pts A student removes 10.00 mL of 0.558M KHP solution and titrates it with a sodium hydroxide solution of unknown concentration. If 13.46mL of NaOH was used to reach the endpoint, what is the concentration of the NaOH solution in units of Molarity? (Be prepared to submit your complete set of calculations.) Type only the numerical portion of your answer in the box below. Your answer will be in units of Molarity (M). Report your answer to 3 significant figures.
The Correct Answer and Explanation is:
Correct answers
- KHP : NaOH mole ratio — 1 : 1
- [NaOH] — 0.415 M
Why the ratio is 1 : 1
Potassium hydrogen phthalate (usually abbreviated KHP) has the formula KHC₈H₄O₄. It is a monoprotic acid; the single acidic hydrogen (the “H” in KHP) can donate one proton. When titrated with sodium hydroxide, that proton is neutralised: KHC8H4O4 (aq)+NaOH (aq) ⟶ KNaC8H4O4 (aq)+H2O (l)\text{KHC}_8\text{H}_4\text{O}_4\;(aq) + \text{NaOH}\;(aq) \;\longrightarrow\; \text{KNaC}_8\text{H}_4\text{O}_4\;(aq) + \text{H}_2\text{O}\;(l)
Because only one OH⁻ ion is required per KHP molecule, the balanced chemical equation shows a 1-to-1 stoichiometric relationship between moles of KHP and moles of NaOH. This is why the mole ratio is 1 mol KHP : 1 mol NaOH.
Determining the NaOH concentration
- Moles of KHP used
nKHP=M×V=0.558 mol L−1×0.01000 L=5.58×10−3 moln_{\text{KHP}} = M \times V = 0.558\;\text{mol L}^{-1} \times 0.01000\;\text{L} = 5.58 \times 10^{-3}\;\text{mol}
- Moles of NaOH required
With the 1 : 1 ratio, the same number of moles of NaOH react:
nNaOH=nKHP=5.58×10−3 moln_{\text{NaOH}} = n_{\text{KHP}} = 5.58 \times 10^{-3}\;\text{mol}
- Volume of NaOH delivered
VNaOH=13.46 mL=0.01346 LV_{\text{NaOH}} = 13.46\;\text{mL} = 0.01346\;\text{L}
- Concentration of NaOH
[NaOH]=nNaOHVNaOH=5.58×10−3 mol0.01346 L=0.41456 mol L−1[\text{NaOH}] = \frac{n_{\text{NaOH}}}{V_{\text{NaOH}}} = \frac{5.58 \times 10^{-3}\;\text{mol}}{0.01346\;\text{L}} = 0.41456\;\text{mol L}^{-1}
Rounded to three significant figures, the molarity is 0.415 M.
Key points to remember
- KHP is monoprotic, so its titration with any strong base is always 1 : 1.
- Convert all volumes to litres before multiplying or dividing by molarity.
- Keep an extra digit during intermediate calculations; round only at the final step to maintain accuracy. This practice ensured the reported 0.415 M value is reliable to three significant figures.
