White vinegar is a \(5.0 \% \) by mass solution of acetic acid in water. If the density of white vinegar is \(1.007\;g/c{m^3}\), what is the \(pH?\)

To calculate the \({\bf{pH}}\), we must first determine the concentration of hydronium ions in the solution. However, we do not know the concentration of the hydronium ions; all we know is the acid's name, density, and mass ratio. Is it feasible to calculate the concentration of the hydronium ions from this information? It certainly is. The formula for calculating the mass ratio is as follows:

\(\begin{aligned}w = \frac{{m( acid )}}{{m( water ) + m( acid )}}\\w = \frac{{m( acid )}}{{m( vinegar )}}.\end{aligned}\)

And the density by:

\(\rho = \frac{{m( vinegar )}}{{V( vinegar )}}.\)

From this formula, we can say that:

\(m (vinegar) = \rho \times V (vinegar).\)

When we add this to the first formula, we get the following:

\(w = \frac{{m (acid)}}{{\rho \times V (vinegar)}}.\)

The acetic acid concentration (molar) can be determined as follows:

\(\begin{aligned}{l}c &= \frac{{n (acid)}}{{V (vinegar)}}\\c &= \frac{{m (acid)}}{{M (acid) \times V (vinegar)}}\\c &= \frac{{m (acid) \times \rho }}{{m (vinegar) \times M (acid)}}.\end{aligned}\)

Since \(\frac{{m (acid)}}{{m (vinegar)}}\) is equal to \(w,\) we can write:

\(c = \frac{{w \times \rho }}{{M (acid)}}.\)

We now have all of the data we need to compute the acetic acid’s content in vinegar. The molar mass of the acetic acid is calculated by adding all of the atomic masses together.

\(\begin{aligned}M\left( {C{H_3}C{O_2}H} \right) &= (2Ar(C) + 4Ar(H) + 2Ar(O))\frac{g}{{mol}}\\M\left( {C{H_3}C{O_2}H} \right) &= (2 \times 12.011 + 4 \times 1.008 + 2 \times 15.999)\frac{g}{{mol}}\\M\left( {C{H_3}C{O_2}H} \right) &= 60.052\frac{g}{{mol}}.\end{aligned}\)

\(\begin{aligned}c &= \frac{{0.05 \times 1.007gcm}}{{60.052gmo{l^{ - 1}}}}\\c &= 8.384 \times 1{0^{ - 4}}\frac{{mol}}{{c{m^3}}}\\c &= 0.838 M.\end{aligned}\)

By multiplying the concentration by \(1000\left( {1L = 1d{m^3} = } \right.1000\;c{m^3})\), we may translate the concentration from \(\frac{{mol}}{{{c^3}}} to \frac{{mol}}{L}.\) We already know how much acetic acid there is, but we also need to know how much hydronium ions there are. Because acetic acid is a weak acid, it does not fully ionize. As a result, we must use the \({K_a}\) value to compute the hydronium ion concentration. The \({K_a}\) value for any weak acid can be found in the table in Appendix \(H\) of the book. The reaction is as follows:

\(C{H_3}C{O_2}H(aq) \to C{H_3}CO_2^ - (aq) + {H^ + }(aq).\)

The \({K_a}\) value is:

\({K_a} = \frac{{c\left( {C{H_3}CO_2^ - } \right) \times c\left( {{H^ + }} \right)}}{{c\left( {C{H_3}C{O_2}H} \right)}}.\)

We can designate the concentrations of \(C{H_3}CO_2^ - \) and \({H^ + }\) as \(x\) and assume that they are equal. We can also assume that the change in the concentration for \(C{H_3}C{O_2}H\) is negligible. We can now write:

\(\begin{aligned}1.8 \times 1{0^{ - 5}} &= \frac{{{x^2}}}{{0.838}} \\x &= c\left( {{H^ + }} \right) &= 3.9 \times 1{0^{ - 3}} M.\end{aligned}\)

Now that we know the concentration of hydronium ions, we can compute the \(pH\) by using the formula:

\(\begin{aligned}pH &= - log\left( {\frac{{c\left( {{H^ + }} \right)}}{{mol{L^{ - 1}}}}} \right)\\pH &= - log\left( {\frac{{3.9 \times 1{0^{ - 3}}mol{L^{ - 1}}}}{{mol{L^{ - 1}}}}} \right)\\pH &= 2.4.\end{aligned}\)