Question

A 1.44 L buffer solution consists of 0.322 M propanoic acid and 0.112 M sodium propanoate. Calculate the pH of the solution following the addition of 0.069 moles of HCl. Assume that any contribution of the HCl to the volume of the solution is negligible. The Ka of propanoic acid is 1.34 × 10-5.

Answer 1

Answer: pH = 4.11

Explanation: pH of the buffer solution is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

pKa is calculated from the given Ka value as:

pKa = - log Ka

$pKa=-log1.34*10^-^5$

pKa = 4.87

pH of the solution before adding HCl to it:

$pH=4.87+log(\frac{0.112}{0.322})$

pH = 4.87 - 0.46

pH = 4.41

Now, 0.069 moles of HCl are added to the buffer solution. This added HCl react with base(sodium propanoate) to produce acid(propanoic acid).

Initial moles of acid = 0.322*1.44 = 0.464

initial moles of base = 0.112*1.44 = 0.161

moles of base after reacting with HCl = 0.161 - 0.069 = 0.092

moles of acid after addition of HCl = 0.464 + 0.069 = 0.533

Let's plug in the values in Handerson equation to calculate the pH:

$pH=4.87+log(\frac{0.092}{0.533})$

pH = 4.87 - 0.76

pH = 4.11

So, the original pH of the buffer solution is 4.41 and after addition of HCl the pH is 4.11 .

Answer 2

The pH of solution following the addition of 0.069 moles of HCl is 4.11.

How to calculate pH?

The pH of a buffer solution can be calculated using Handerson equation as follows:

pH = pka + log (base/acid)

pKa of the acid is calculated from the given Ka value as follows:

pKa = - log Ka

pKa = - log 1.34 × 10-⁵

pKa = 4.87

The pH of the solution before adding HCl to it is as follows:

pH = 4.87 + log(0.112/0.322)

pH = 4.87 - 0.46 = 4.41

According to this question, 0.069 moles of HCl are added to the buffer solution.

• Initial moles of acid = 0.322 × 1.44 = 0.464
• Initial moles of base = 0.112 × 1.44 = 0.161

• moles of base after reacting with HCl = 0.161 - 0.069 = 0.092
• moles of acid after addition of HCl = 0.464 + 0.069 = 0.533

Therefore, the pH of the buffer after adding HCl is:

pH = 4.87 + log(0.092/0.533)

pH = 4.87 - 0.76 = 4.11

Therefore, the pH of solution following the addition of 0.069 moles of HCl is 4.11.

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