Question

answer the following questions based on the reaction below: NaOH(aq) + KHP(s) --> NaKP(aq)+H2O(I)

A 1.864 g sample of impure KHP was titrated with a 0.0978 M solution of NaOH. To completely react the KHP in the sample, 72.14 mL of base was needed. KHP (potassium hydrogen phthalate, 204.23 g/mol)

a.) How many grams of KHP were in the unknown sample?

b.) What is the percentage of KHP in the unknown sample?

Answer 1

Answer:

• a) 1.44g

• b) 77.3%

Explanation:

1. Chemical balanced equation (given)

       $NaOH(aq)+ KHP(s)\rightarrow NaKP(aq)+H_2O(l)$

2. Mole ratio

$1molNaOH(aq):1molKHP(s)$

This is, 1 mol of NaOH will reacts with 1 mol of KHP.

3. Find the number of moles in 72.14 mL of the base

    $Molarity=\text{number of moles of solute}/\text{volume of solution in liters}$

    $\text{Volume of solution}=72.14mL=0.07214liters$

     $\text{Number of moles of NaOH}=0.0978M\times 0.07214liter=0.007055mol$

4. Find the number of grams of KHP that reacted

The number of moles of KHP that reacted is equal to the number of moles of NaOH, 0.007055 mol

Convert moles to grams:

• mass = number moles × molar mass = 0.007055mol × 204.23g/mol

• mass = 1.4408 g.

You have to round to 3 significant figures: 1.44 g (because the molarity is given with 3 significant figures).

5. Find the percentage of KHP in the sample

The percentage is how much of the substance is in 100 parts of the sample.

The formula is:

• % = (mass of substance / mass of sample) × 100

• % = (1.4408g/ 1.864g) × 100 = 77.3%