A flask of 0.25 M HBrO(aq) has what pH? (pK_a = 8.69)

Question

Quizplus · Accepted Answer

First, we need to write the equilibrium reaction for HBrO dissociating in water:
HBrO (aq) + H2O (l) ⇌ H3O+ (aq) + BrO- (aq)

The Ka for this reaction can be calculated by:
Ka = [H3O+][BrO-]/[HBrO]

We know the concentration of HBrO is 0.25 M, so we just need to find the concentrations of H3O+ and BrO- at equilibrium. Since HBrO is a weak acid, we can assume that the dissociation is small, and that the concentration of HBrO at equilibrium is equal to its initial concentration. Therefore, we can simplify the Ka expression to:
Ka = [H3O+][BrO-]/0.25

Since HBrO and BrO- are a conjugate acid-base pair, we can use the pKa equation to find the concentration of H3O+:
pKa = -log(Ka)
pKa = -log([H3O+][BrO-]/0.25)
pKa = -log([H3O+]) - log([BrO-]) + log(0.25)
8.69 = -log([H3O+]) - log([BrO-]) + log(0.25)
log([H3O+]) = -8.69 + log([BrO-]) - log(0.25)
log([H3O+]) = -5.172 - log([BrO-])
[H3O+] = 10^(-5.172 - log([BrO-]))

At equilibrium, the concentration of H3O+ is equal to the concentration of BrO-. We can use the fact that the sum of the concentrations of HBrO, H3O+, and BrO- is equal to the initial concentration of HBrO (0.25 M) to set up an ICE table:

HBrO    H3O+    BrO-
I (initial)   0.25    0       0
C (change)    -x      +x      +x
E (equilibrium)0.25-x  x       x

Substituting the equilibrium values into the Ka expression:
Ka = [H3O+][BrO-]/0.25
Ka = x^2/0.25
x = √(0.25Ka)

Now we can find the pH using the concentration of H3O+:
pH = -log([H3O+])
pH = -log(10^(-5.172 - log([BrO-])))
pH = -(−5.172 − log([BrO-])) = 5.172 + log([BrO-])
pH = 5.172 + log(√(0.25Ka))
pH = 5.172 + log(√(0.25*10^(-8.69)))
pH = 4.65

Therefore, the pH of the 0.25 M HBrO solution is approximately 4.65, which is closest to answer choice D.

A Flask of 0