Question

Consider 1.0 L of an aqueous solution that contains 0.10 M sulfuric acid to which 0.30 mole barium nitrate is added. Assuming no change in volume of the solution, determine the pH, the concentration of barium ions in the final solution, and the mass of solid formed.

Short answer

The pH is $0.70$, the concentration of barium ions in the final solution is$0.20\mathrm{M}$ and the mass of solid formed is $23.34\mathrm{g}$.

Step-by-step solution

Introduction to the concept.

The number of moles of material that may be dissolved per liter of solution before the solution becomes saturated is known as molar solubility.

Determination of the pH, the concentration, and the mass of produced.

From the given, the prior species present in the given solution are, ${\mathbf{H}}^{\mathbf{+}}\mathbf{,}{\mathbf{HSO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{,}{\mathbf{Ba}}^{\mathbf{2}\mathbf{+}}\mathbf{,}{\mathbf{NO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{,}\mathbf{and}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

The disassociation of${\mathrm{HSO}}_4^{-}$ produces by${\mathrm{Ba}}^{2+}$ reacts with ${\mathrm{SO}}_4^{2+}$.

${\mathrm{HSO}}_4^{-}\left(\mathrm{aq}\right)\to {\text{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)$

${\mathrm{K}}_{\mathrm{a}2}=1.2\times {\text{10}}^{-2}$

${\mathrm{Ba}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{-}\left(\mathrm{aq}\right)\to {\mathrm{BaSO}}_4\left(\mathrm{s}\right)$

$\mathrm{K}=6.7\times {10}^8$

By adding above two equations,

${\mathrm{Ba}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{HSO}}_4^{-}\left(\mathrm{aq}\right)\to {\text{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{BaSO}}_4\left(\mathrm{s}\right)$

${\mathrm{K}}_{\mathrm{overall}}=8.0\times {10}^6$

Here,${\mathrm{K}}_{\mathrm{overall}}$is so large. Let’s complete the reaction. We know that $H_{2}S{O}_4$ is the strongest.

${\left[{\mathrm{HSO}}_4^{-}\right]}_0={\left[{\mathrm{H}}^{+}\right]}_0=0.10\text{M}$

Initial change	$0.30\text{M-}0.10\text{M}$	$0.10\text{M-}0.10\text{M}$	$0.10\text{M+}0.10\text{M}$
After change	$0.20\text{M}+x$	$0+x$	$0.20\text{M-}x$
eqm	$\left(0.20+x\right)$	$x$	$\left(0.20-x\right)$

The reaction's equilibrium constant will be:

$\begin{array}{c}\mathrm{K}=\frac{\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{Ba}}^{2+}\right]\left[{\mathrm{HSO}}_4^{-}\right]}\\ 8.0\times {10}^6=\frac{0.20-\mathrm{x}}{\left(0.20+\mathrm{x}\right)\mathrm{x}}\approx \frac{0.20}{0.20\mathrm{x}}\end{array}$

$\mathrm{x}=1.3\times {10}^{-7}\text{M}$

The pH of barium ions in the final solution.

$\left[{\mathrm{H}}^{+}\right]=0.20-1.3\times {10}^{-7}\text{M}$

$\mathrm{pH}=-\log \left(0.20\right)$

$\mathrm{pH}=0.70$

The concentration of barium ions in the final solution.

$\left[{\mathrm{Ba}}^{2+}\right]=0.20+1.3\times {10}^{-7}\text{M}$

$=0.20\text{M}$

Thus, the number of moles of${\mathrm{BaSO}}_4$ produced $=0.10\text{mol}$.