Question

You have weighed out 1.0312 g of KHP mw (204.24 g/mol) on the analytical balance and dissolved it into 25.00 mL of DI water. You then diluted the KHP solution to 250.00 mL.

You took a 50.00 mL aliquot of this diluted sample and titrated it with an NaOH solution to the equivalence point. 20.47 mL of NaOH was used for the titration.

The neutralization reaction that occurred was a 1:1 ratio of KHP to NaOH.

What is the molarity of the NaOH? correct sig figs.

Answer 1

$$C_1V_1 = C_2V_2$$

$C_1 = C(KHP) = n(KHP)/V(KHP)$

$$n(KHP) = m/M = 1.03129g/204.24(g/mol) = 0.00505mol$$

$$C(KHP) = 0.00505mol/0.250L = 0.0202M$$

$$0.0202*50 = C(NaOH)*20.47$$

$$C(NaOH)= 0.04934M$$