Calculate the percent ionization of cyanic acid, K_a = 2.0 × 10^-4, in a buffer solution that is 0.50 M HCNO and 0.10 M NaCNO.

Question

Quizplus · Accepted Answer

HCNO is a weak acid, so it will produce H+ and CNO- ions in solution. NaCNO is a salt of the conjugate base of HCNO, so it will provide CNO- ions. The buffer solution will have a pH close to the pKa of HCNO, which is 3.48.

To calculate the percent ionization of HCNO, we need to determine the concentrations of the acid and the conjugate base ([HCNO] and [CNO-], respectively) and the concentration of undissociated HCNO ([H+] is negligible, since the pH is close to the pKa).

From the given concentrations of 0.50 M HCNO and 0.10 M NaCNO, we can assume that the initial concentration of CNO- is negligible compared to the amount produced by the dissociation of HCNO. Therefore, we can treat [CNO-] as equal to the concentration of HCNO that ionizes.

Let x be the extent of ionization of HCNO. Then:

HCNO ⇌ H+ + CNO-
Initial:      0.50 M          0             0
Change:     -x               +x           +x
Equilibrium: 0.50 - x        x            x

The equilibrium constant for this reaction is Ka = [H+][CNO-] / [HCNO] = x^2 / (0.50 - x).

Since x is small compared to 0.50, we can approximate (0.50 - x) as 0.50. Then:

Ka = x^2 / 0.50
x = sqrt(Ka * 0.50)

For HCNO, Ka = 1.2 x 10^-4, so:

x = sqrt(1.2 x 10^-4 * 0.50) = 0.012

The percent ionization is:

% ionization = (moles of ionized HCNO / initial moles of HCNO) x 100%
= (0.012 mol / 0.50 mol) x 100% = 2.4%.

However, we need to account for the effect of the added salt (NaCNO) on the ionization of HCNO. The presence of a common ion (CNO-) will shift the equilibrium to the left, reducing the ionization of HCNO.

We can use the common ion effect equation:

Ka' = Ka / (1 + [CNO-] / Ka)

To calculate the new equilibrium constant and ionization extent.

Since [CNO-] = 0.10 M and Ka = 1.2 x 10^-4:

Ka' = 1.2 x 10^-4 / (1 + 0.10 M / 1.2 x 10^-4) = 8.0 x 10^-6

x' = sqrt(Ka' * 0.50) = 0.006

The percent ionization with the added salt is:

% ionization = (moles of ionized HCNO / initial moles of HCNO) x 100%
= (0.006 mol / 0.50 mol) x 100% = 1.2%.

Therefore, the answer is (C) 0.20%.

Calculate the Percent Ionization of Cyanic Acid, Ka = 2

Calculate the Percent Ionization of Cyanic Acid, K_a = 2