We´re asked to **calculate the pH** of a **0.1 M Ca(CH _{3}COO)_{2} solution** at 25 °C.

Calcium acetate is an ionic salt soluble in water.

In it, Ca(CH_{3}COO)_{2} dissociates as follows:

**Ca(CH _{3}COO)_{2}**

The expression of the equilibrium constant **K** is:

$\overline{){\mathbf{K}}{\mathbf{=}}\frac{\mathbf{p}\mathbf{r}\mathbf{o}\mathbf{d}\mathbf{u}\mathbf{c}\mathbf{t}\mathbf{s}}{\mathbf{r}\mathbf{e}\mathbf{a}\mathbf{c}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{t}\mathbf{s}}}$

In solution, the acetate ion can react with water by accepting a proton to form acetic acid:

**CH _{3}CO_{2}^{-}_{(aq)} + H_{2}O_{(l)} **

Since acetate ion accepts a proton, it is a base according to the Bronsted-Lowry theory.

For a weak base, the equilibrium constant becomes the basic dissociation constant (K_{b}).

The expression for K_{b} is:

$\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}\frac{\left[\mathbf{AH}\right]\left[{\mathbf{OH}}^{\mathbf{-}}\right]}{\left[{\mathbf{A}}^{\mathbf{-}}\right]}}$

For the calcium acetate ion reaction, K_{b} adopts the form of:

$\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}\frac{\left[\mathbf{C}{\mathbf{H}}_{\mathbf{3}}\mathbf{C}\mathbf{O}\mathbf{O}\mathbf{H}\right]\left[\mathbf{O}{\mathbf{H}}^{\mathbf{-}}\right]}{\left[\mathbf{C}{\mathbf{H}}_{\mathbf{3}}\mathbf{C}{\mathbf{O}}_{\mathbf{2}}^{\mathbf{-}}\right]}}$

Note that each concentration is raised by the stoichiometric coefficient: [CH_{3}COOH], [HO^{-}], and [CH_{3}CO_{2}^{-}] are raised to 1.

Calcium acetate (Ca(CH_{3}COO)_{2}) is commonly used as a food additive, especially in candy. What is the pH of a 0.1 M Ca(CH_{3}COO)_{2} solution at 25 °C

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