What is the hydronium ion concentration in a solution prepared by mixing 50.00 mL of 0.10 M HCN with 50.00 mL of 0.030 M NaCN? Assume that the volumes of the solutions are additive and that
K_a = 4.9 × 10^-10 for HCN.

Question

Quizplus · Accepted Answer

First, write the balanced chemical equation to determine the reaction between HCN and NaCN:

HCN + NaCN → HCN(aq) + CN-(aq) + Na+(aq)

Since HCN is a weak acid, it will partially dissociate in water, forming hydronium ions:

HCN(aq) + H2O(l) ⇌ H3O+(aq) + CN-(aq)

Using the given equilibrium constant (Ka) of HCN:

Ka = [H3O+][CN-] / [HCN]

rearrange to solve for [H3O+]:

[H3O+] = Ka * [HCN] / [CN-]

Since the volumes of the solutions are additive, the total moles of HCN and CN- can be calculated.

moles HCN = 0.10 M x 0.050 L = 0.005 mol
moles CN- = 0.030 M x 0.050 L = 0.0015 mol

The [HCN] and [CN-] in the equation can now be expressed as:

[HCN] = 0.005 mol / 0.100 L = 0.050 M
[CN-] = 0.0015 mol / 0.100 L = 0.015 M

Plugging in these values and the given Ka into the equation for [H3O+]:

[H3O+] = (4.9x10^-10)(0.050 M) / (0.015 M) = 1.633x10^-9 M

Therefore, the hydronium ion concentration is 1.6x10^-9 M, which corresponds to answer choice C.

What Is the Hydronium Ion Concentration in a Solution Prepared