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

3.07 × 10⁻⁴

Step-by-step 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 * 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 = (5.22 * 10^(-3) mol )/(0.088L) = 0.0593M`

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

`Ka = ([H^(+)]^(2) )/(Ca) = ((4.27 * 10^(-3) )^(2) )/(0.0593) = 3.07 * 10^(-4)`