From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for ea

step 1 At equilibrium, the value of equilibrium constant is equal to the reaction quotient for the reac

From the equilibrium concentrations given, calculate Ka for...

From the equilibrium concentrations given, calculate $K_{\mathrm{a}}$ for each of the weak acids and $K_{\mathrm{b}}$ for each of the weak bases. $$ \begin{array}{l}{\text { (a) } \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} :\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.34 \times 10^{-3} \mathrm{M}} \\ {\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]=1.34 \times 10^{-3} \mathrm{M}} \\ {\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=9.866 \times 10^{-2} \mathrm{M}}\end{array} $$ $$ \begin{array}{l}{\text { (b) } \mathrm{ClO}^{-} :\left[\mathrm{OH}^{-}\right]=4.0 \times 10^{-4} \mathrm{M}} \\ {[\mathrm{HClO}]=2.38 \times 10^{-5} \mathrm{M}} \\ {\left[\mathrm{ClO}^{-}\right]=0.273 \mathrm{M}}\end{array} $$ $$ \begin{array}{l}{\text { (c) } \mathrm{HCO}_{2} \mathrm{H} :\left[\mathrm{HCO}_{2} \mathrm{H}\right]=0.524 M} \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=9.8 \times 10^{-3} M} \\ {\left[\mathrm{HCO}_{2}-\right]=9.8 \times 10^{-3} \mathrm{M}}\end{array} $$ $$ \begin{array}{l}{\text { (d) } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} :\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right]=0.233 \mathrm{M} ;} \\ {\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right]=2.3 \times 10^{-3} \mathrm{M}} \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=2.3 \times 10^{-3} \mathrm{M}}\end{array} $$

From the equilibrium concentrations given, calculate $K_{\mathrm{a}}$ for each of the weak acids and $K_{\mathrm{b}}$ for each of the weak bases.
$$
\begin{array}{l}{\text { (a) } \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} :\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.34 \times 10^{-3} \mathrm{M}} \\ {\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]=1.34 \times 10^{-3} \mathrm{M}} \\ {\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]=9.866 \times 10^{-2} \mathrm{M}}\end{array}
$$
$$
\begin{array}{l}{\text { (b) } \mathrm{ClO}^{-} :\left[\mathrm{OH}^{-}\right]=4.0 \times 10^{-4} \mathrm{M}} \\ {[\mathrm{HClO}]=2.38 \times 10^{-5} \mathrm{M}} \\ {\left[\mathrm{ClO}^{-}\right]=0.273 \mathrm{M}}\end{array}
$$
$$
\begin{array}{l}{\text { (c) } \mathrm{HCO}_{2} \mathrm{H} :\left[\mathrm{HCO}_{2} \mathrm{H}\right]=0.524 M} \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=9.8 \times 10^{-3} M} \\ {\left[\mathrm{HCO}_{2}-\right]=9.8 \times 10^{-3} \mathrm{M}}\end{array}
$$
$$
\begin{array}{l}{\text { (d) } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} :\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right]=0.233 \mathrm{M} ;} \\ {\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right]=2.3 \times 10^{-3} \mathrm{M}} \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=2.3 \times 10^{-3} \mathrm{M}}\end{array}
$$

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