Calculating the grams of KHP needed for NaOH + HCl titration

Question

In my titration lab, the [HCl] is trying to be determined by first using KHP to standardize NaOH. I'm given two scenarios: either [NaOH] = 10[HCl] or 10[NaOH] = [HCl], where the concentration of NaOH is around 0.1M or 1M.

If a 100.0 mL solution of KHP is prepared to react with 25.00 mL of NaOH, how much KHP has to be weighed out in each scenario?

Answer 1

First, I'll assume that KHP is potassium hydrogen phthalate and not potassium hydrogen phosphate. That being the case, we have the following reaction:

NaOH + KHP ==> NaKP + H2O (typical acid base neutralization reaction)

If [NaOH] = 0.1 M:

25 ml x 1 L/1000 ml x 0.1 mol/L = 0.0025 moles NaOH in 25 ml

Since the stoichiometric ratio of KHP : NaOH is 1:1, you would thus need 0.0025 moles KHP

0.0025 moles KHP x 204.2 g/mol = 0.5105 grams of KHP

If [NaOH] = 1.0 M:

25 ml x 1 L/1000 ml x 1.0 mol/L = 0.025 moles NaOH in 25 ml

Since the stoichiometric ratio of KHP : NaOH is 1:1, you would thus need 0.025 moles KHP

0.025 moles KHP x 204.2 g/mol = 5.105 grams KHP