Problem 90 - Chemistry

Question

What are the major species present in the following mixtures of bases? a. \(0.050 \mathrm{M} \mathrm{NaOH}\) and \(0.050 \mathrm{M} \mathrm{LiOH}\) b. \(0.0010 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) and \(0.020 \mathrm{M} \mathrm{RbOH}\) What is \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of each of these solutions?

Step-by-Step Solution

Answer

In the first solution with NaOH and LiOH, the major species present are Na\(^+\), Li\(^+\), and OH\(^-\). The concentration of OH\(^-\) ions is 0.100 M, and the pH of the solution is 13. In the second solution with Ca(OH)\(_2\) and RbOH, the major species present are Ca\(^{2+}\), Rb\(^+\), and OH\(^-\). The concentration of OH\(^-\) ions is 0.022 M, and the pH of the solution is approximately 12.34.

Step 1: a. Identify the major species in the solution of \(\mathrm{NaOH}\) and \(\mathrm{LiOH}\)

First, we need to identify the major species present in the solution. These are the ions that are produced when the strong bases, \(\mathrm{NaOH}\) and \(\mathrm{LiOH}\), dissociate completely in water: \(\mathrm{NaOH \rightarrow Na^+ + OH^-}\) \(\mathrm{LiOH \rightarrow Li^+ + OH^-}\) So, the major species present in the solution are \(\mathrm{Na^+}\), \(\mathrm{Li^+}\), and \(\mathrm{OH^-}\) ions.

Step 2: b. Calculate the concentration of \(\mathrm{OH}^-\) ions

Since both \(\mathrm{NaOH}\) and \(\mathrm{LiOH}\) are strong bases, they will dissociate completely in water. Thus, the total concentration of \(\mathrm{OH^-}\) ions will be the sum of the concentrations contributed by both bases. \(\left[\mathrm{OH}^-\right] = 0.050\,\mathrm{M} + 0.050\,\mathrm{M} = 0.100\,\mathrm{M}\)

Step 3: c. Calculate the \(\mathrm{pH}\)

First, we need to calculate the pOH using the relation: \(\mathrm{pOH} = - \log_{10} \left[\mathrm{OH}^-\right]\) \(\mathrm{pOH} = - \log_{10} (0.100) = 1\) Now we can use the relationship between \(\mathrm{pH}\) and \(\mathrm{pOH}\): \(\mathrm{pH} + \mathrm{pOH} = 14\) \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 1 = 13\) The pH of the solution is 13.

Step 4: a. Identify the major species in the solution of \(\mathrm{Ca(OH)_2}\) and \(\mathrm{RbOH}\)

Similarly, we need to identify the major species present in the solution of \(\mathrm{Ca(OH)_2}\) and \(\mathrm{RbOH}\). These bases also dissociate completely in water: \(\mathrm{Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^-}\) \(\mathrm{RbOH \rightarrow Rb^+ + OH^-}\) So, the major species present in the solution are \(\mathrm{Ca^{2+}}\), \(\mathrm{Rb^+}\), and \(\mathrm{OH^-}\) ions.

Step 5: b. Calculate the concentration of \(\mathrm{OH}^-\) ions

First, let's find the contribution of \(\mathrm{OH}^-\) ions from each base: From \(\mathrm{Ca(OH)_2}\): \(0.0010\,\mathrm{M} \times 2 = 0.0020\,\mathrm{M}\) From \(\mathrm{RbOH}\): \(0.020\,\mathrm{M}\) Now, we add both contributions: \(\left[\mathrm{OH}^-\right] = 0.0020\,\mathrm{M} + 0.020\,\mathrm{M} = 0.022\,\mathrm{M}\)

Step 6: c. Calculate the \(\mathrm{pH}\)

First, we need to calculate the pOH using the relation: \(\mathrm{pOH} = - \log_{10} \left[\mathrm{OH}^-\right]\) \(\mathrm{pOH} = - \log_{10} (0.022) \approx 1.66\) Now we can use the relationship between \(\mathrm{pH}\) and \(\mathrm{pOH}\): \(\mathrm{pH} + \mathrm{pOH} = 14\) \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 1.66 \approx 12.34\) The pH of the solution is approximately 12.34.