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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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.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Weak Acids concept. If you need more Weak Acids practice, you can also practice Weak Acids practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Vasquez's class at OKSTATE.