Question

A 5.22 × 10−3−mol sample of HY is dissolved in enough H2O to form 0.088 L of solution. If the pH of the solution is 2.37, what is the Ka of HY?

Answer 1

Answer:

3.07 × 10⁻⁴

Explanation:

Step 1: Calculate the concentration of H⁺

We will use the definition of pH.

$pH = -log [H^{+} ]\\\[ [H^{+} ] = antilog -pH = antilog -2.37 = 4.27 \times 10^{-3} M$

Step 2: Calculate the concentration of HY

5.22 × 10⁻³ mol of HY are dissolved in 0.088 L. The concentration of the acid (Ca) is:

$Ca = \frac{5.22 \times 10^{-3} mol }{0.088L} = 0.0593M$

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

$Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(4.27 \times 10^{-3} )^{2} }{0.0593} = 3.07 \times 10^{-4}$