Problem: Calculate the pH of a solution prepared from 0.201 mol of NH4CN and enough water to make 1.00 L of solution.

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FREE Expert Solution

We’re being asked to calculate the pH of a solution 0.201 mol NH4CN and enough water to make 1.00 L of solution.


NH4CN dissociates in water in the following manner:

NH4CN(s) ⇌ NH4+ (aq) + CN- (aq)


NH4+ is the conjugate acid of a weak base, NH3:

NH3 + H2O ⇌ NH4+ + OH-Kb=1.76x10-5


CN- is the conjugate base of a weak acid, HCN:

HCN + H2O ⇌ CN- + H3OKa= 4.9x10-10


The pH of the salt of a weak acid and a weak base can be calculated using the following equation: 

pH = 12(pKw + pKa-pKb)



We will do the following steps to solve this problem:

Step 1: Calculate pKa of the weak acid

Step 2: Calculate the pKof the weak base

Step 3: Determine pKw

Step 4: Calculate the pH of the salt

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Problem Details

Calculate the pH of a solution prepared from 0.201 mol of NH4CN and enough water to make 1.00 L of solution.