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A sample of 0.1687 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.1150 M NaOH. The acid req uired 15.5 mL of base to reach the equivalence point. What is the molar mass of the acid?

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(a)

HA(aq)+NaOH(aq)NaA(aq)+H2O(l)\mathrm{HA(aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)}

The 25.0 mL of monoprotic acid solution contains 0.1687 g HA

The acid requires 15.5 mL of 0.1150 M NaOH to reach the equivalence point

Let us calculate the molar mass of the acid.

First, let us calculate the number of moles of NaOH

nNaOH=0.0155L0.1150M=0.00178mol\begin{align*} n_{NaOH} &= 0.0155 \mathrm{L} \cdot 0.1150 \mathrm{M}\\ &= 0.00178 \mathrm{mol} \end{align*}

To reach the equivalence point, one mole of NaOH will react with one mole of HA, therefore, the number of moles of monoprotic acid is also 0.00178 mol.

Now, let us find the molar mass of monoprotic acid

MHA=mHAnHA=0.1687g0.00178mol=94.775g/mol\begin{align*} M_{HA} &= \frac { m_{HA} } { n_{HA} }\\ &= \frac { 0.1687 \mathrm{g} } { 0.00178 \mathrm{mol} }\\ &= {\color{#4257b2}94.775 \mathrm{g/mol}} \end{align*}

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