Potassium hydrogen phthalate is a monoprotic acid. Dihydrogen phthalate (H2C8H4O4) is a diprotic acid. An analyte sample contains 0.127 M KHP and 0.0678 M H2C8H4O4. What volume of a 0.205 M NaOH solution is required to neutralize 25.0 mL of the analyte solution?
Molar masses:
H2C8H4O4 - 166.13 g/mol
KHP - 204.23 g/mol
I think something is missing but I don't know what.
KHP + NaOH ==> H2O + NaKP
H2P + 2NaOH ==> 2H2O + Na2P
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Suppose the sample is 100% KHP, then
25.0 mL x 0.127M = 3.175 mmols.
That will take 3.175 mmols NaOH
M NaOH = mmols/mL; rearrange to
mL = mmols NaOH/M NaOH= 3.175/0.205 = about 15 mL of the NaOH. That's one possible solution.
Suppose the sample is 100% H2P, then
25.0 mL x 0.0678M = 1.695 mmols H2P.
mmols NaOH = 2 x 1.695 = 3.39 and
3.39/0.205 = about 16 mL of the NaOH. That's a second possible solution.
Doesn't it stand to reason that for any possible combination there will be a solution; therefore, there is no unique solution.
To determine the volume of the NaOH solution required to neutralize the analyte solution, we need to understand the stoichiometry of the reaction between the acids (KHP and H2C8H4O4) and the base (NaOH). Let's break down the process step by step:
1) Write the balanced equation for the reaction:
KHP + NaOH → KNaP + H2O
2) Determine the stoichiometry of the reaction:
From the equation, we can see that 1 mole of KHP reacts with 1 mole of NaOH to produce 1 mole of KNaP and 1 mole of water.
3) Calculate the moles of KHP and H2C8H4O4 in the analyte solution:
Moles of KHP = concentration (0.127 M) × volume (25.0 mL) / 1000
Moles of KHP = 0.127 × 25.0 / 1000 = 0.003175 mol
Moles of H2C8H4O4 = concentration (0.0678 M) × volume (25.0 mL) / 1000
Moles of H2C8H4O4 = 0.0678 × 25.0 / 1000 = 0.001695 mol
4) Determine the limiting reagent:
Since the reaction requires an equal number of moles for both KHP and NaOH and the moles of KHP (0.003175 mol) > moles of H2C8H4O4 (0.001695 mol), KHP is the limiting reagent.
5) Calculate the moles of NaOH required to react with the limiting reagent (KHP):
Moles of NaOH = moles of KHP = 0.003175 mol
6) Determine the volume of the NaOH solution:
Molarity of NaOH = 0.205 M
Moles of NaOH = 0.003175 mol
Volume of NaOH = moles of NaOH / Molarity of NaOH
Volume of NaOH = 0.003175 / 0.205 = 0.0155 L
Convert the volume to mL:
Volume of NaOH = 0.0155 × 1000 = 15.5 mL
Therefore, 15.5 mL of a 0.205 M NaOH solution is required to neutralize 25.0 mL of the analyte solution.