If the pKa of lactic acid is 5, then the pH of 0.005 M calcium lactate solution at 25°C is _____ $$\times 10^{-1}$$ (Nearest integer)

Calcium lactate, $$\text{Ca(C}_3\text{H}_5\text{O}_3)_2$$, is a salt formed from a weak acid (lactic acid) and a strong base ($$\text{Ca(OH)}_2$$).

When dissolved in water, it dissociates completely:

$$\text{Ca(Lactate)}_2 \rightarrow \text{Ca}^{2+} + 2\,\text{Lactate}^-$$

Determine the Concentration of the Lactate Ion

Since each mole of calcium lactate produces $$2$$ moles of lactate ions ($$\text{Lactate}^-$$), the total concentration of the basic anion ($$C$$) is:

$$C = 2 \times 0.005\text{ M} = 0.01\text{ M} = 10^{-2}\text{ M}$$

Calculate the pH

The pH of a salt of a weak acid and a strong base undergoing anionic hydrolysis is calculated using the formula:

$$\text{pH} = 7 + \frac{1}{2}(\text{p}K_a + \log C)$$

Given:

• $$\text{p}K_a = 5$$
• $$\log C = \log(10^{-2}) = -2$$

Substitute these values into the equation:

$$\text{pH} = 7 + \frac{1}{2}(5 - 2)$$
$$\text{pH} = 7 + \frac{1}{2}(3)$$
$$\text{pH} = 7 + 1.5 = 8.5$$

$$\text{pH} = \text{Value} \times 10^{-1}$$

Setting $$8.5 = X \times 10^{-1}$$, we get:

$$X = 85$$