Question

Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions. It has the unwieldy formula of KHC8H4O4. This is often written in shorthand notation as KHP. What volume of a 0.327 M barium hydroxide solution is needed to exactly neutralize 2.75 grams of KHP

Answer 1

Answer: The volume of barium hydroxide needed is 20.64 mL

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of KHP = 2.75 g

Molar mass of KHP = 204.22 g/mol

Putting values in above equation, we get:

$\text{Moles of KHP}=\frac{2.75g}{204.22g/mol}=0.0135mol$

The chemical reaction for the reaction of KHP and barium hydroxide follows:

$2KHC_8H_4O_4(aq.)+Ba(OH)_2\rightarrow Ba(KC_8H_4O_4)_2(aq.)+2H_2O(l)$

By Stoichiometry of the reaction:

2 moles of KHP reacts with 1 mole of barium hydroxide

So, 0.0135 moles of KHP will react with = $\frac{1}{2}\times 0.0135=0.00675mol$ of barium hydroxide.

To calculate the molarity of barium hydroxide, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Moles of barium hydroxide = 0.00675 moles

Molarity of solution = 0.327 M

Putting values in above equation, we get:

$0.327=\frac{0.00675mol\times 1000}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=\frac{0.00675\times 1000}{0.327}=20.64mL$

Hence, the volume of barium hydroxide needed is 20.64 mL