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What is the ph of a 0.45 m solution of NaCN?

User Cadolphs
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Final answer:

The pH of a 0.45 M solution of NaCN is 13.65.

Step-by-step explanation:

The pH of a solution can be calculated using the formula:

pH = -log[H+]

Given that NaCN is a salt of a weak acid (HCN), it will undergo hydrolysis in water:

HCN + H2O ↔ H3O+ + CN-

The equilibrium can be represented by the expression:

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

Since NaCN is a strong electrolyte, it will fully dissociate into Na+ and CN- ions:

NaCN → Na+ + CN-

This means that the concentration of CN- ions will be equal to the concentration of NaCN. Therefore, the pH of a 0.45 M NaCN solution will depend on the dissociation of CN- ions:

NaCN → Na+ + CN-

[CN-] = 0.45 M

pOH = -log[CN-] = -log(0.45) = 0.35

pH = 14 - pOH = 14 - 0.35 = 13.65.

User Apollo
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