A 3.555 g sample of a monoprotic acid was dissolved in water. It took 23.32 mL of a 0.875 M NaOH to neutralize the acid. Calculate the molar mass of the acid.
mols NaOH = M x L = ?
mols HA = mols NaOH
mols HA = grams HA/molar mass HA
Solve for molar mass.
To calculate the molar mass of the monoprotic acid, we can use the concept of stoichiometry, which relates the balanced equation of the reaction to the number of moles involved.
First, we need to determine the number of moles of sodium hydroxide (NaOH) used to neutralize the acid. We can use the equation:
Moles of NaOH = Molarity of NaOH x Volume of NaOH
Moles of NaOH = (0.875 mol/L) x (23.32 mL / 1000 mL/L) [Convert mL to L]
Moles of NaOH = 0.875 x 0.02332 mol
Moles of NaOH = 0.02029 mol
Next, since the reaction between the acid and NaOH is 1:1, we know that the number of moles of the acid is also 0.02029 mol.
Now, we can calculate the molar mass of the acid using the formula:
Molar mass = Mass of acid / Moles of acid
Given that the mass of the acid is 3.555 g, we can substitute it into the equation:
Molar mass = 3.555 g / 0.02029 mol
Molar mass = 175.1 g/mol
Therefore, the molar mass of the monoprotic acid is 175.1 g/mol.