Question

Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.0550 M$\mathbf{Ba}{\left(\mathrm{OH}\right)}_2$ solution. What is the concentration of the excess ${\mathbf{H}}^{+}$ or ${\mathbf{OH}}^{-}$left in the solution?

Short answer

The concentration of excess$\left[O{H}^{-}\right]$ left in the solution is 0.02 M.

Step-by-step solution

The balanced chemical equation

The balanced chemical equation of the reaction is:

$Ba{\left(OH\right)}_2\left(aq\right)+2HCl\left(aq\right)\to BaC{l}_2\left(aq\right)+2H_{2}O\left(l\right)$

One mole of $Ba{\left(OH\right)}_2$reacts with 2 mol of HCl.

The number of moles of each constituent

The number of moles of$HCl$ is:

$\begin{array}{c}{n}_{HCl}=M_{HCl}\times V_{HCl}\\ =0.250M\times 0.075L\\ =0.01875mol\end{array}$

The number of moles of$Ba{\left(OH\right)}_2$ is:

$\begin{array}{c}{n}_{Ba{\left(OH\right)}_2}=M_{Ba{\left(OH\right)}_2}\times V_{Ba{\left(OH\right)}_2}\\ =0.0550M\times 0.225L\\ =0.012375mol\end{array}$

The difference in the number of moles

From the balanced chemical equation,

2 mol of HCl react with 1 mol of$Ba{\left(OH\right)}_2$

Then, 0.01875 mol of HCl will react with the following number of moles of $Ba{\left(OH\right)}_2$:

$\begin{array}{c}{n}_{Ba{\left(OH\right)}_2}=\frac{0.01875mol}{2}\\ =0.009375mol\end{array}$

The unreacted number of moles of $Ba{\left(OH\right)}_2$is:

$\begin{array}{c}{n}_{unreacted}=0.012375mol-0.009375mol\\ =0.003mol\end{array}$

The concentration of $Ba{\left(OH\right)}_2$that remains unreacted is:

$\begin{array}{c}\left[Ba{\left(OH\right)}_2\right]=\frac{n_{Ba{\left(OH\right)}_2}}{V_{total}}\\ =\frac{0.003mol}{\left(0.225+0.075\right)L}\\ =0.01M\end{array}$

The excess hydroxide concentration in the solution is:

$\begin{array}{c}\left[O{H}^{-}\right]=2\times 0.01M\\ =0.02M\end{array}$