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Strong base is dissolved in 665 ml. of 0.200 M weak acid (K, 4.73 x 10) to make a buffer with a pH of 4.10. Assume that the volume remains constant when the base is added.

HA(aq) + OH (aq) → H₂O(I)+A(aq)

Calculate the pK, value of the acid, and determine the number of moles of acid initially present. Using pH= pKa + log((A-)/HA))

1 answer

To calculate the \( pK_a \) value of the weak acid and determine the number of moles of acid initially present, we will follow these steps:

  1. Calculate \( pK_a \) from \( K_a \).
  2. Determine the moles of the weak acid (HA) initially present using the concentration and volume given.
  3. Use the Henderson-Hasselbalch equation to find the number of moles of its conjugate base (A\(^-\)).

Step 1: Calculate \( pK_a \)

The dissociation constant \( K_a \) for the weak acid is given as \( 4.73 \times 10^{-5} \).

The \( pK_a \) can be calculated using the formula: \[ pK_a = -\log(K_a) \]

Calculating: \[ pK_a = -\log(4.73 \times 10^{-5}) \approx 4.33 \]

Step 2: Determine Initial Moles of Acid (HA)

The concentration of the weak acid (HA) is \( 0.200 , M \), and the volume is \( 665 , mL \): \[ \text{Volume in liters} = 665 , mL \times \frac{1 , L}{1000 , mL} = 0.665 , L \]

Calculating moles of the weak acid (HA): \[ \text{Moles of HA} = \text{Concentration} \times \text{Volume} = 0.200 , M \times 0.665 , L = 0.133 , \text{moles} \]

Step 3: Use the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is:

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

We know:

  • pH = 4.10
  • \( pK_a \) = 4.33
  • Initial moles of HA = 0.133 moles

Rearranging the equation to find the ratio \(\frac{[A^-]}{[HA]}\):

\[ 4.10 = 4.33 + \log \left(\frac{[A^-]}{[HA]}\right) \] \[ \log \left(\frac{[A^-]}{[HA]}\right) = 4.10 - 4.33 = -0.23 \] \[ \frac{[A^-]}{[HA]} = 10^{-0.23} \approx 0.585 \]

Finding Moles of Conjugate Base (A^-)

Let \( x \) be the moles of base added, then at equilibrium:

  • Moles of HA remaining = \( 0.133 - x \)
  • Moles of A⁻ formed = \( x \)

Using the ratio we found: \[ \frac{x}{0.133 - x} = 0.585 \] Solving for x:

  1. Cross-multiplying: \[ x = 0.585(0.133 - x) \] \[ x = 0.077805 - 0.585x \]
  2. Collecting terms: \[ x + 0.585x = 0.077805 \] \[ 1.585x = 0.077805 \] \[ x \approx \frac{0.077805}{1.585} \approx 0.049 \text{ moles of } A^- \]

Summary

  • The \( pK_a \) of the weak acid is approximately 4.33.
  • The number of moles of the weak acid (HA) initially present is 0.133 moles.

The calculations show how the weak acid and its conjugate base interact in making a buffer solution at a specific pH.