Questions: A solution is prepared at 25°C that is initially 0.86 M in propanoic acid (HC2H5CO2), a weak acid with Ka=1.3 x 10^-5, and 1.2 M in potassium propanoate (KC2H5CO2). Calculate the pH of the solution. Round your answer to 2 decimal places. pH=

A solution is prepared at 25°C that is initially 0.86 M in propanoic acid (HC2H5CO2), a weak acid with Ka=1.3 x 10^-5, and 1.2 M in potassium propanoate (KC2H5CO2). Calculate the pH of the solution. Round your answer to 2 decimal places.

pH=

Transcript text: A solution is prepared at $25^{\circ} \mathrm{C}$ that is initially 0.86 M in propanoic acid $\left(\mathrm{HC}_{2} \mathrm{H}_{5} \mathrm{CO}_{2}\right)$, a weak acid with $K_{a}=1.3 \times 10^{-5}$, and 1.2 M in potassium propanoate $\left(\mathrm{KC}_{2} \mathrm{H}_{5} \mathrm{CO}_{2}\right)$. Calculate the pH of the solution. Round your answer to 2 decimal places. \[ \mathrm{pH}= \]

Solution

Solution Steps

Step 1: Identify the Type of Solution

The solution is a buffer solution because it contains a weak acid (propanoic acid) and its conjugate base (potassium propanoate).

Step 2: Use the Henderson-Hasselbalch Equation

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: \[ \mathrm{pH} = \mathrm{pK_a} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \] where $[\text{A}^-]$ is the concentration of the conjugate base and $[\text{HA}]$ is the concentration of the weak acid.

Step 3: Calculate $\mathrm{pK_a}$

Calculate $\mathrm{pK_a}$ from the given $K_a$: \[ \mathrm{pK_a} = -\log(K_a) = -\log(1.3 \times 10^{-5}) \]

Step 4: Substitute Values into the Henderson-Hasselbalch Equation

Substitute the values into the equation: \[ \mathrm{pH} = \mathrm{pK_a} + \log \left( \frac{1.2}{0.86} \right) \]

Step 5: Calculate the pH

Perform the calculations to find the pH and round to 2 decimal places.