A 3.664 g sample of a monoprotic acid was dissolved in water. It took 20.27 mL of a 0.1578 M NaOH solution to neutralize the acid. Calculate the molar mass of the acid. (Monoprotic refers to an acid that provides only one H+).

Molar mass of monoprotic acid is 1145g/mol.

Explanation:

The reaction of a monoprotic acid (HX) with NaOH is:

HX + NaOH → H₂O + NaX

That means 1 mole of acid reacts with 1 mole of NaOH

If the neutralization of the acid spent 20.27mL of a 0.1578 M NaOH solution. Moles of NaOH are:

0.02027L × (0.1578mol / L) = 3.199x10⁻³ moles NaOH ≡ moles HX.

As mass of the sample is 3.664g, molar mass of the acid is:

3.664g / 3.199x10⁻³ moles = 1145g/mol

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