Calculate either [H3O+] or [OH−] for each of the solutions at 25 °C. Solution A: [OH−]=3.29×10−7 M; [H3O+]= _M Solution B: [H3O+]=9.63×10−9 M; [OH−]= M Solution C: [H3O+]=6.49×10−4 M; [OH−]= __M Which of these solutions are basic at 25 °C? Solution A: [OH−]=3.29×10−7 M Solution C: [H3O+]=6.49×10−4 M Solution B: [H3O+]=9.63×10−9 M

[tex]\begin{gathered} solution\text{ A: }[H_3O^+]=3.039\times10^{-8}M \\ solution\text{ B: }[OH^-]=1.038\times10^{-6}M \\ solution\text{ C: }[OH^-]=1.54\times10^{-11}M \end{gathered}[/tex]

Explanations:

Using the formula below to determine the concentration of the hydroxonium ion;

[tex]\begin{gathered} [H_3O^+][OH^-]=1\times10^{-14} \\ [H_3O^+]=\frac{1\times10^{-14}}{3.29\times10^{-7}} \\ [H_3O^+]=\frac{1}{3.29}\times10^{-14-(-7)} \\ [H_3O^+]=0.3039\times10^{^{-7}} \\ [H_3O^+]=3.039\times10^{-8}M \end{gathered}[/tex]

Determine the pH

[tex]\begin{gathered} pH=-log[H_3O] \\ pH=-log(3.039\times10^{-8}) \\ pH=-(-7.5) \\ pH=7.5 \end{gathered}[/tex]

This shows that the solution A is BASIC since the pH is greater than 7

For solution B

[tex]\begin{gathered} [OH^-]=\frac{1\times10^{-14}}{[H_3O^+]} \\ [OH^-]=\frac{1\times10^{-14}}{9.63\times10^{-9}} \\ [OH^-]=0.1038\times10^{-5}M \\ [OH^-]=1.038\times10^{-6}M \end{gathered}[/tex]

Determine the pH

[tex]\begin{gathered} pH=-log[H_3O^+] \\ pH=-log(9.63\times10^{-9}) \\ pH=-(-8.02) \\ pH=8.02 \end{gathered}[/tex]

This shows that the solution B is BASIC since the pH is greater than 7

For the solution C

[tex]\begin{gathered} [OH^-]=\frac{1\times10^{-14}}{[H_3O^+]} \\ [OH^-]=\frac{1\times10^{-14}}{6.49\times10^{-4}} \\ [OH^-]=0.1540\times10^{-10}M \\ [OH^-]=1.54\times10^{-11}M \end{gathered}[/tex]

Determine the pH of the solution C

[tex]\begin{gathered} pH=-log[H_3O^+] \\ pH=-log(6.49\times10^{-4}) \\ pH=-(-3.19) \\ pH=3.19 \end{gathered}[/tex]

Since the pH of the soution C is less than 7, hence the solution C is ACIDIC