What quantity in moles of NaOH need to be added to 200.0 mL of a 0.200 M solution of HF to make a buffer with a pH of 4.00? (Ka for HF is 6.8 × 10⁻⁴)

To solve this problem, we need to use the Henderson-Hasselbalch equation, which is:

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

Where:

• pH is the negative log of the H+ concentration
• pKa is the negative log of the Ka
• [A-] is the concentration of the base (NaOH in this case)
• [HA] is the concentration of the acid (HF in this case)

Step 1: Calculate pKa from Ka pKa = -log(Ka) = -log(6.8 × 10⁻⁴) = 3.17

Step 2: Substitute the known values into the Henderson-Hasselbalch equation 4.00 = 3.17 + log([NaOH]/[HF])

Step 3: Solve for [NaOH]/[HF] log([NaOH]/[HF]) = 4.00 - 3.17 = 0.83 [NaOH]/[HF] = 10^0.83 = 6.76

Step 4: Solve for [NaOH] We know that the initial concentration of HF is 0.200 M, so: [NaOH] = 6.76 * [HF] = 6.76 * 0.200 M = 1.35 M

Step 5: Convert the concentration of NaOH to moles We know that the volume of the solution is 200.0 mL = 0.200 L, so: moles of NaOH = [NaOH] * volume = 1.35 M * 0.200 L = 0.27 moles

Therefore, 0.27 moles of NaOH need to be added to the solution to make a buffer with a pH of 4.00.