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What is the pH of a 0.440 M solution of Ca(NO₂)₂ (Ka of HNO₂ is 4.5 × 10⁻⁴)?

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Final answer:

The pH of a 0.440 M solution of Ca(NO₂)₂ can be calculated by first finding the concentration of H3O+ ions. Since Ca(NO₂)₂ is a salt, it dissociates completely in water. Using the dissociation reaction of HNO₂, we can relate the concentration of H3O+ to the concentration of NO₂-. From there, we can calculate the concentration of H3O+ and find the pH.

Step-by-step explanation:

To calculate the pH of a solution, we need to determine the concentration of the hydronium ion (H3O+). In this case, the solution is a 0.440 M solution of Ca(NO₂)₂, but we need to find the concentration of H3O+. Since Ca(NO₂)₂ is a salt, it dissociates completely in water. So, we need to determine the concentration of NO₂- and use that to find the concentration of H3O+ using the dissociation reaction of HNO₂:

HNO₂ (aq) + H₂O (l) → H₃O+ (aq) + NO₂- (aq)

The Ka value of HNO₂ is given as 4.5 × 10⁻⁴. Using the equilibrium expression for the dissociation of HNO₂, we can relate the concentration of H3O+ to the concentration of NO₂-:

Ka = [H3O+][NO₂-] / [HNO₂]

We can assume that the concentration of HNO₂ remains the same as the initial concentration, which is 0.440 M. Therefore, we can rearrange the equation to find the concentration of H3O+:

[H3O+] = Ka * [HNO₂] / [NO₂-]

[H3O+] = (4.5 × 10⁻⁴) * (0.440) / (0.440)

[H3O+] ≈ 4.5 × 10⁻⁴ M

To find the pH, we can take the negative logarithm of the concentration of H3O+:

pH = -log[H3O+]

pH = -log(4.5 × 10⁻⁴)

pH ≈ 3.35

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