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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+).

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Answer:

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

Eduard22sly

The molar mass of the monoprotic acid, HX is 1145 g/mol

We'll begin by calculating the number of mole of NaOH in the solution. This can be obtained as follow:

Volume = 20.27 mL = 20.27 / 1000 = 0.02027 L

Molarity of NaOH = 0.1578 M

Mole of NaOH =?

Mole = Molarity x Volume

Mole of NaOH = 0.1578 × 0.02027

Mole of NaOH = 0.0032 mole

  • Next, we shall determine the number of mole of the monoprotic acid, HX required to react with 0.0032 mole of NaOH. This can be obtained as follow

HX + NaOH → H₂O + NaX

From the balanced equation above,

1 mole of NaOH reacted with 1 mole of HX.

Therefore,

0.0032 mole of NaOH will also react with 0.0032 mole of HX.

  • Finally, we shall determine the molar mass of 0.0032 mole of HX. This can be obtained as follow:

Mole of HX = 0.0032 mole

Mass of HX = 3.664 g

Molar mass of HX =?

Molar mass = mass / mole

Molar mass of HX = 3.664 / 0.0032

Molar mass of HX = 1145 g/mol

Thus, the molar mass of the monoprotic acid, HX is 1145 g/mol

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