pH of a Weak Acid SolutionCalculate the pH of a 1.28 mol.L–1 solution of hydrofluoric acid. The pKa of HF is 3.17.Enter your answer to 2 decimal places.Answer:

To calculate the pH of a weak acid solution, we can use the formula for the ionization of a weak acid HA ⇌ H+ + A- and the definition of the acid dissociation constant Ka.

Step 1: Write the ionization equation of the acid. HF ⇌ H+ + F-

Step 2: Write the expression for the acid dissociation constant Ka. Ka = [H+][F-]/[HF]

Step 3: Since the degree of ionization is small, the concentration of HF at equilibrium will be approximately equal to the initial concentration. So, [HF] ≈ 1.28 M.

Step 4: Let x be the concentration of H+ ions at equilibrium. Since HF is a monoprotic acid, [H+] = [F-] = x.

Step 5: Substitute these values into the Ka expression. Ka = x^2 / (1.28 - x)

Step 6: Since HF is a weak acid, we can assume that x << 1.28. So, the Ka expression simplifies to: Ka = x^2 / 1.28

Step 7: Solve this equation for x. But first, we need to convert pKa to Ka. Ka = 10^-pKa = 10^-3.17 = 0.000677

0.000677 = x^2 / 1.28 x^2 = 0.000677 * 1.28 x = sqrt(0.000677 * 1.28) x = 0.0295 M

Step 8: The pH of the solution is given by the negative logarithm (base 10) of the H+ ion concentration. pH = -log[H+] pH = -log(0.0295) pH = 1.53

So, the pH of a 1.28 M solution of hydrofluoric acid is approximately 1.53.