What is the percentage of KHP in the mixture?

Question

Suppose you are titrating a solid mixture containing KHP (molar mass=204.2 g/mol) and a salt compound. KHP reacts with NaOH according to the equation:

KHP+NaOH ⟶ H2O + NaKP

*You prepare the analyte from 0.505 g of the KHP mixture. You then require 11.95 mL of 0.118 M NaOH solution to complete the titration. What is the percentage of KHP in the mixture?

Answer 1

@Annona D,

Since you've received 2 answers that do not agree with each other, I thoiught I'd weigh in and see what answer I get.

KHP + NaOH ==> H₂O + NaKP

moles NaOH used = 11.95 ml x 1 L / 1000 ml x 0.118 mols / L = 0.001410 mols NaOH used

moles KHP present = 0.001410 mols NaOH x 1 mol KHP / mol NaOH = 0.001410 mols KHP

mass KHP present = 0.001410 mols KHP x 204.2 g kHP / mol = 0.2879 g KHP present

% by mass KHP = 0.2879 g / 0.505 g (x100%) = 57.0% (3 sig. figs.)

Answer 2

Dear student,

First, it is essential to underline the role of KHP in the titration described above. Since KHP reacts with NaOH, the total KHP within the mixture decreases as the titration progresses and NaKP is produced.

titration occurs => moles of KHP decrease

We can first think of the critical moments of the titration to formulate our answer.

1. Before the titration occurs, we have an initial amount of KHP present in the mixture; let such amount of KHP be denoted by KHP_initial.
2. After the titration occurs, some of the KHP in the mixture will not react. The KHP that did not react with the NaOH of our titration will be denoted by KHP_final.

The percentage of KHP in the mixture after the titration occurs then becomes:

( KHP_final / KHP_initial )*100 = % KHP in mixture after titration

Since the question already provides the mass of KHP_initial (0.505 g in analyte), we must find KHP_final. Considering that we needed 11.95 mL of 0.118 M of NaOH to complete the titration, we can determine the amount of NaOH that reacted and, eventually, KHP_final.

11.95 mL NaOH * (1 L solution / 1000 mL solution) * (0.118 mol NaOH / 1 L solution) =

0.0014101 mol NaOH reacted

As indicated by the given chemical equation, one mole of NaOH reacts with one mole of KHP; thus, the amount of KHP that reacted is also equal to 0.0014101 moles. Using the molecular weight of KHP (204.2 g/mol), we can conclude that about 0.28794 g of KHP reacted and became NaKP.

Since we are calculating the percentage of KHP in the mixture after the titration, our percentage then becomes:

( initial KHP- reacted KHP) / ( initial KHP) * 100 = ( final KHP) / ( initial KHP) * 100

(0.505 g - 0.2879 g ) / ( 0.505 g) *100 = (0.2171 g) / ( 0.505 g)*100 = 43 %

Answer 3

11.95 mL NaOH x 1L NaOH x 0.118 moles NaOH x 1 mole KHP x 204.2g KHP = 0.288g KHP

1000mL NaOH 1L NaOH 1 moles NaOH 1 mole KHP

mass percent of KHP = 0.288g KHP x 100

0.505g KHP

= 57% KHP