Question

Determine the molar mass of an unknown monoprotic acid to two decimal places if 16.98 mL of a 0.086 M NaOH solution were used to titrate 0.236 g of the unknown acid

Answer 1

1 ) Chemical equation

`\text{NaOH + HX}\rightarrow H_2O+Na^++X^-`

HX represents the unknown acid.

2) Moles of NaOH in the reaction

`M=\frac{\text{moles of solute}}{\text{liters of solution}}`

Convert mL into L

`L=16.98mL\cdot(1L)/(1000mL)=0.01698L`

Plug in known values in the equation and solve for moles.

`0.086M=\frac{\text{moles of NaOH}}{0.01698L}`
`\text{mol NaOH= 0.086M}\cdot0.01698L=0.00146028\text{ mol NaOH}`

3) Moles of the unknown acid that reacted with 0.00146028 mol NaOH

Molar ratio

1 mol NaOH: 1 mol HX

`\text{mol HX=0.00146028 mol NaOH}\cdot\frac{1\text{ mol HX}}{1\text{ mol NaOH}}=0.00146028\text{ mol HX}`

4) Molar mass of the unknown monoprotic acid

`\text{Molar Mass=}\frac{\text{mass of solute (g)}}{moles\text{ of solute}}`

Plug in known values and solve

`\text{Molar Mass}=\frac{0.236\text{ g HX}}{0.00146028\text{ mol HX}}=161.61\text{ g/mol}`

The molar mass of the unknown monoprotic acid is 161.61 g/mol

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