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A sample of a monoprotic acid (HA) weighing 0.384 g is dissolved in water and the solution is titrated with aqueous NaOH. If 30.0 mL of 0.100 M NaOH is required to reach the equivalence point, what is the molar mass of HA?

(a) 211 g/mol
(b) 128 g/mol
(c) 81.0 g/mol
(d) 37.0 g/mol
(e) 20.3 g/mol

Respuesta :

Answer:

The correct answer is option b.

Explanation:

Moles of NaOH = n

Volume of NaOH solution = 30.0 mL =0.030 L ( 1mL =0.001 L)

Molarity of the NaOH solution = 0.100 M

[tex]Molarity=\frac{Moles}{Volume(L)}[/tex]

[tex]n=0.100 M\times 0.030 L=0.00300 mol[/tex]

[tex]HA(aq)+NaOH(aq)\rightarrow NaA(aq)+H_2O(aq)[/tex]

According to reaction , 1 mol of NaOH reacts with 1 mole of HA, then, 0.00300 moles of NaOH will react with :

[tex]\frac{1}{1}\times 0.00300 mol=0.00300 mol[/tex] of HA

Moles of HA = 0.00300 mol

Mass of HA = 0.384 g

Molar mass of HA = M

Mass of 0.00300 moles of HA = 0.384 g

[tex]0.00300 mol\times M=0.384 g[/tex]

[tex]M=\frac{0.384 g}{0.00300 mol}=128 g/mol[/tex]

128 g/mol is the molar mass of HA.

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