Questions: A chemistry graduate student is given 450 mL of a 1.20 M chlorous acid (HClO2) solution. Chlorous acid is a weak acid with Ka=1.1 x 10^-2. What mass of KClO2 should the student dissolve in the HClO2 solution to turn it into a buffer with pH=2.51? You may assume that the volume of the solution doesn't change when the KClO2 is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.

A chemistry graduate student is given 450 mL of a 1.20 M chlorous acid (HClO2) solution. Chlorous acid is a weak acid with Ka=1.1 x 10^-2. What mass of KClO2 should the student dissolve in the HClO2 solution to turn it into a buffer with pH=2.51?

You may assume that the volume of the solution doesn't change when the KClO2 is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.

Transcript text: A chemistry graduate student is given $450 . \mathrm{mL}$ of a 1.20 M chlorous acid $\left(\mathrm{HClO}_{2}\right)$ solution. Chlorous acid is a weak acid with $K_{a}=1.1 \times 10^{-2}$. What mass of $\mathrm{KClO}_{2}$ should the student dissolve in the $\mathrm{HClO}_{2}$ solution to turn it into a buffer with $\mathrm{pH}=2.51$ ? You may assume that the volume of the solution doesn't change when the $\mathrm{KClO}_{2}$ is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.

Solution

Solution Steps

Step 1: Determine the Concentration of Hydronium Ions

To find the concentration of hydronium ions $[\text{H}^+]$ in the buffer solution, we use the pH formula:

\[ \text{pH} = -\log[\text{H}^+] \]

Given that $\text{pH} = 2.51$, we can solve for $[\text{H}^+]$:

\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-2.51} \approx 3.089 \times 10^{-3} \, \text{M} \]

Step 2: Use the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation for a buffer solution is:

\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]

Where:

$[\text{A}^-]$ is the concentration of the conjugate base $\text{ClO}_2^-$.
$[\text{HA}]$ is the concentration of the weak acid $\text{HClO}_2$.
$\text{pK}_a = -\log K_a$.

Calculate $\text{pK}_a$:

\[ \text{pK}_a = -\log(1.1 \times 10^{-2}) \approx 1.959 \]

Substitute the known values into the Henderson-Hasselbalch equation:

\[ 2.51 = 1.959 + \log\left(\frac{[\text{ClO}_2^-]}{1.20}\right) \]

Step 3: Solve for the Concentration of $[\text{ClO}_2^-]$

Rearrange the equation to solve for $[\text{ClO}_2^-]$:

\[ 2.51 - 1.959 = \log\left(\frac{[\text{ClO}_2^-]}{1.20}\right) \]

\[ 0.551 = \log\left(\frac{[\text{ClO}_2^-]}{1.20}\right) \]

Convert from logarithmic form to exponential form:

\[ 10^{0.551} = \frac{[\text{ClO}_2^-]}{1.20} \]

\[ [\text{ClO}_2^-] = 1.20 \times 10^{0.551} \approx 3.794 \, \text{M} \]

Step 4: Calculate the Mass of $\text{KClO}_2$ Required

The concentration of $\text{ClO}_2^-$ is the same as the concentration of $\text{KClO}_2$ because they dissociate completely. Calculate the moles of $\text{KClO}_2$ needed:

\[ \text{Moles of } \text{KClO}_2 = 3.794 \, \text{M} \times 0.450 \, \text{L} = 1.7073 \, \text{mol} \]

Calculate the mass of $\text{KClO}_2$ using its molar mass ($M_{\text{KClO}_2} = 106.55 \, \text{g/mol}$):

\[ \text{Mass of } \text{KClO}_2 = 1.7073 \, \text{mol} \times 106.55 \, \text{g/mol} \approx 182.0 \, \text{g} \]