🤓 Based on our data, we think this question is relevant for Professor Vasquez's class at OKSTATE.
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- + H3O+ ; Ka= 4.9x10-10
The pH of the salt of a weak acid and a weak base can be calculated using the following equation:
We will do the following steps to solve this problem:
Step 1: Calculate pKa of the weak acid
Step 2: Calculate the pKb of the weak base
Step 3: Determine pKw
Step 4: Calculate the pH of the salt
Calculate the pH of a solution prepared from 0.201 mol of NH4CN and enough water to make 1.00 L of solution.