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

Short Answer

Expert verified
The major species present in the mixtures are: a. \(\mathrm{Na}^+\), \(\mathrm{Li}^+\), \(\mathrm{OH}^-\) b. \(\mathrm{Ca}^{2+}\), \(\mathrm{Rb}^+\), \(\mathrm{OH}^-\) The hydroxide ion concentrations are: a. \(\left[\mathrm{OH}^-\right] = 0.100\ \mathrm{M}\) b. \(\left[\mathrm{OH}^-\right] = 0.022\ \mathrm{M}\) The pH values of the mixtures are: a. pH ≈ 13 b. pH ≈ 12.34

Step by step solution

01

Find the major species in each mixture

In each mixture, we have to consider the dissociation of each base in water. The base dissociation reaction for a general base is given by: \[ B + H_2O \rightarrow BH^+ + OH^- \] We can apply this reaction for both the mixtures separately to find the major species present. a. Mixture with 0.050 M NaOH and 0.050 M LiOH \[ \mathrm{NaOH} \rightarrow \mathrm{Na}^+ + \mathrm{OH}^- \] \[ \mathrm{LiOH} \rightarrow \mathrm{Li}^+ + \mathrm{OH}^- \] Major species present: \(\mathrm{Na}^+\), \(\mathrm{Li}^+\), \(\mathrm{OH}^-\) b. Mixture with 0.0010 M Ca(OH)2 and 0.020 M RbOH \[ \mathrm{Ca(OH)_2} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^- \] \[ \mathrm{RbOH} \rightarrow \mathrm{Rb}^+ + \mathrm{OH}^- \] Major species present: \(\mathrm{Ca}^{2+}\), \(\mathrm{Rb}^+\), \(\mathrm{OH}^-\)
02

Find the hydroxide ion concentration \(\left[ OH^- \right]\) for each mixture

Now, we need to find the total concentration of hydroxide ions for each mixture by summing up the contributions from each base. a. For the first mixture, Total \(\left[\mathrm{OH}^-\right] = [\mathrm{OH}^-]_{\mathrm{NaOH}} + [\mathrm{OH}^-]_{\mathrm{LiOH}} = 0.050 + 0.050 = 0.100\ \mathrm{M}\) b. For the second mixture, Total \(\left[\mathrm{OH}^-\right] = [\mathrm{OH}^-]_{\mathrm{Ca(OH)_2}} + [\mathrm{OH}^-]_{\mathrm{RbOH}} = 2 (0.0010) + 0.020 = 0.022\ \mathrm{M}\)
03

Calculate the pOH and pH values for each mixture

Now, we can use the relationship between pOH and \(\left[ \mathrm{OH}^- \right]\) as well as the relationship between pH and pOH to find the pH of each mixture. The formula for pOH is: \[ \mathrm{pOH} = -\log_{10}[\mathrm{OH}^-] \] The relationship between pH and pOH is: \[ \mathrm{pH} + \mathrm{pOH} = 14 \] a. For the first mixture, pOH = \(-\log_{10}(0.100) = 1\) pH = \(14 - 1 = 13\) b. For the second mixture, pOH = \(-\log_{10}(0.022) \approx 1.66\) pH = \(14 - 1.66 \approx 12.34\) So, the pH values for the mixtures are approximately 13 and 12.34, respectively.

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