Question

A solution is prepared at 25∘C that is initially 0.084M in propanoic acid (HC2H5CO2 ), a weak acid with Ka =1.3×10 ^−5 , and 0.50M in sodium propanoate (NaC2H5CO2 ). Calculate the pH of the solution. Retain at least two decimal places when rounding your answer

Answer 1

The pH of a solution prepared with propanoic acid and sodium propanoate can be calculated using the ICE approach. The initial pH of the solution is approximately 2.87.

Step-by-step explanation:

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

pH = -log[H3O+]

In this case, the solution is prepared with propanoic acid and sodium propanoate. The initial pH of the solution can be calculated using the ICE approach and the given values of the acid concentration and Ka:

[H3O+] = Ka × [CH3CO2H] = √(1.3 × 10^-5 × 0.084) = 2.87 × 10^-3

Therefore, the pH of the solution is approximately 2.87.

Answer 2

The pH of the buffer solution consisting of propanoic acid and sodium propanoate is calculated using the Henderson-Hasselbalch equation. By inserting the given values into the equation, we determine the pH to be 5.665, indicating a slightly acidic buffer solution.

Step-by-step explanation:

To calculate the pH of the solution containing propanoic acid and its conjugate base, sodium propanoate, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid. Given that the concentration of propanoic acid (HA) is 0.084M and the concentration of sodium propanoate (A-) is 0.50M, and the pKa is the negative logarithm of Ka (pKa = -log(Ka)), we can substitute the values into the equation (pKa for propanoic acid is 4.89, calculated from Ka = 1.3 x 10-5):

pH = -log(1.3 x 10-5) + log(0.50/0.084)

Which gives us:

pH = 4.89 + log(5.952)

Now calculate the log(5.952):

pH = 4.89 + 0.775

The final pH of the solution is:

pH = 5.665

This value indicates the slightly acidic nature of the solution due to the weak acid propanoic acid and the presence of its conjugate base, which stabilizes the pH.