Question

A 0.375 g sample of a monoprotic acid is dissolved in water and titrated with 0.270 M KOH. What is the molar mass of the acid if 11.0 mL of the KOH solution is required to neutralize the sample?

Answer 1

To find the molar mass of the monoprotic acid, use the given volume and concentration of KOH to calculate the moles of KOH used. Then, since the acid and KOH react in a 1:1 ratio, the molar mass of the acid can be calculated using its mass and the number of moles.

Step-by-step explanation:

To find the molar mass of the monoprotic acid, we first need to determine the number of moles of KOH used in the titration. The balanced chemical equation for the reaction is:

H3PO4 (aq) + 3 KOH (aq) -> K3PO4 (aq) + 3 H2O (l)

From the equation, we can see that 1 mole of H3PO4 reacts with 3 moles of KOH. Using the given volume of KOH solution (11.0 mL) and its concentration (0.270 M), we can calculate the number of moles of KOH used:

Moles of KOH = volume (L) x concentration (M) = 0.0110 L x 0.270 M = 0.00297 mol KOH

Since the acid and KOH react in a 1:1 ratio, the number of moles of the acid is equal to the number of moles of KOH. Therefore, the molar mass of the acid can be calculated using its mass (0.375 g) and the number of moles:

Molar mass = mass (g) / moles = 0.375 g / 0.00297 mol = 126.26 g/mol