Answer to Question #136282 in Chemistry for Miranda

Question

a quantity of 14.68 mL of a KOH solution is needed to neutralize o.4018 g of KHP. what is the concentration (in molarity) of the KOH solution? (the molar mass of KHP is 204.2 g/moL.)

Answer 1

The balanced reaction equation between the weak acid potassium hydrogen phthalate or KHP (C₈H₅O₄K) and potassium hydroxide is as follows:

KOH + C₈H₅O₄K $\rightarrow$ C₈H₄O₄K₂ + H₂O.

As one can see from the stoichiometry of the equation, 1 mol of KOH reacts with 1 mol of KHP:

$n(KOH) = n(KHP)$ .

The number of the moles of KHP can be calculated from its molar mass $M$ = 204.22 g/mol and its mass $m$ :

$n = \frac{m}{M} = \frac{0.4018 \text{ g}}{204.22 \text{ g/mol}} = 0.001967$ mol.

Therefore, the number of the moles of KOH, used for the reaction, is:

$n(KOH) = 0.001967$ mol.

The concentration of the solution is the number of the moles of the solute (KOH) divided by the volume of the solution:

$c = \frac{0.001967\text{ mol}}{14.68·10^{-3}\text{ L}} = 0.1340$ mol/L.

Answer: the concentration (in molarity) of the KOH solution is 0.1340 mol/L.