Question

Determine \(\left[\mathrm{OH}^{-}\right], \mathrm{pOH}\) and \(\mathrm{pH}\) of a \(0.28 \mathrm{M}\) aqueous solution of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\)

Short answer

Answer: The pH of a 0.28 M aqueous solution of Na₂CO₃ is approximately 13.448.

Step-by-step solution

Write the balanced chemical equation

\(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is a salt that dissociates completely in water. First, it breaks down into sodium ions (\(\mathrm{Na}^+\)) and carbonate ions (\(\mathrm{CO}_3^{2-}\)): $$\mathrm{Na}_{2}\mathrm{CO}_{3}\,\mathrm{(aq)} \rightarrow 2\,\mathrm{Na}^{+}\,\mathrm{(aq)} + \mathrm{CO}_3^{2-}\,\mathrm{(aq)}$$ Since carbonate ions are a weak base, they will react with water to produce hydroxide ions (\(\mathrm{OH}^-\)) and bicarbonate ions (\(\mathrm{HCO}_3^-\)): $$\mathrm{CO}_3^{2-}\,\mathrm{(aq)} + \mathrm{H}_2\mathrm{O} \leftrightharpoons \mathrm{HCO}_3^-\,\mathrm{(aq)} + \mathrm{OH}^-\,\mathrm{(aq)}$$

Calculate the concentration of hydroxide ions (\(\left[\mathrm{OH}^{-}\right]\))

Given that the initial concentration of \(\mathrm{Na}_2\mathrm{CO}_3\) is \(0.28\,\mathrm{M}\) and it dissociates completely, the initial concentration of carbonate ions (\(\mathrm{CO}_3^{2-}\)) will also be \(0.28\,\mathrm{M}\). Since the equilibrium constant (\(K_b\)) for the second reaction is small, we can assume that all \(\mathrm{CO}_3^{2-}\) ions react to produce \(\mathrm{OH}^-\) ions. Therefore, we can directly use the initial concentration of carbonate ions as the concentration of hydroxide ions: $$\left[\mathrm{OH}^{-}\right] = 0.28\,\mathrm{M}$$

Calculate the pOH

Now that we have the concentration of hydroxide ions, we can calculate the pOH using the formula: $$\mathrm{pOH} = -\log \left[\mathrm{OH}^{-}\right]$$ Plug in the value of the hydroxide ion concentration: $$\mathrm{pOH} = -\log (0.28) \approx 0.552$$

Calculate the pH

Finally, we can calculate the pH using the relationship between pH and pOH: $$\mathrm{pH} + \mathrm{pOH} = 14$$ Plug in the value of pOH and solve for pH: $$\mathrm{pH} = 14 - 0.552 \approx 13.448$$ In conclusion, for a \(0.28\,\mathrm{M}\) aqueous solution of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), the concentration of hydroxide ions (\(\left[\mathrm{OH}^{-}\right]\)) is \(0.28\,\mathrm{M}\), the pOH is approximately \(0.552\), and the pH is approximately \(13.448\).