Question

Calculate the pH of a solution that is prepared by dissolving 0.740 mol of hypochlorous acid (HCIO, Ka = 3.00x10-8) and 0.254 mol of hydrofluoric acid (HF, K, = 6.60*10-4) in water and diluting to 4.50 L. Also, calculate the equilibrium concentrations of HCIO, CIO-, HF, and F. Do not make an approximation unless the initial acid concentration is greater than 1000 ~ Kg. (Hint: The pH will be determined by the stronger acid of this pair.) pH = [HCIO] = [CIO-]= [HF] = [F]=

Answer 1

For HClO;

Initial [HClO] = moles/volume = 0.740/4.50 = 0.1644 M

1000 * Ka = 1000 * 3*10^-8 = 3*10^-5

Thus, approximation can be made here as

0.1644 > 3*10^-5

HClO -------> H+ + ClO- , Ka = 3*10^-8

0.1644 0 0 (at t =0)

0.1644-x x x (at equilibrium)

Ka = [H+] [ClO-] / [HClO]

3*10^-8 = x * x / (0.1644 - x)

Since, Ka is very less then x is very less.

So, 0.1644 - x = 0.1644

3*10^-8 * 0.1644 = x^2

x = 7.024*10^-5

Thus, at equlibrium;

[H+ ] = [ClO-] = x = 7.024*10^-5 M

[HClO] = 0.1644 - x = 0.1644 M

For HF;

Initial [HF] = 0.254/4.50 = 0.05644 M

1000 * Ka = 1000 * 6.60 * 10^-4 = 0.66

0.66 > 0.05644

Hence approximation cannot be made.

HF ------> H+ + F- , Ka = 6.60*10^-4

0.05644 0 0 (at t=0)

0.05644-x x x (at equilibrium)

6.60*10^-4 = x*x / (0.05644-x)

0.00003725 - 0.00066x = x^2

x^2 + 0.00066x - 0.00003725 = 0

After solving above quadratic equation;

x = 0.005782

Hence at equilibrium;

[H+] = [F-] = x = 0.005782 M

[HF] = 0.05644-x = 0.05644 - 0.005782 = 0.05066 M

Since HF is stronger acid thus [H+] is higher for it from HClO (7.024*10^-5 - very less concentration). And [H+] from HF will drive pH of solution.

pH = - log [H+] = - log (0.005782) = 2.24

[HClO] = 0.1644 M

[ClO-] = 7.024*10^-5 M

[HF] = 0.05066 M

[F-] = 0.005782 M

Answer