Question

Consider the titration of a 50 mL sample of 0.180 M HF with 0.210 M NaOH. Determine each quantity. The Ka for hydrofluoric acid is 3.5 x 10⁻⁴. Calculate the volume of added base that is required to reach the equivalence point. (Hint: what equals what at the equivalence point?)

a) 25 mL
b) 50 mL
c) 75 mL
d) 100 mL
The initial pH of HF. Calculate the pH after adding 5.0 mL of NaOH. (Hint: What kind of solution have you made? What equation can you use to calculate the pH?)

a) 1.74
b) 2.48
c) 3.22
d) 4.01

Answer 1

The volume of added base required to reach the equivalence point in the titration of HF with NaOH is 50 mL. The initial pH of the HF solution is approximately 3.46. The pH after adding 5.0 mL of NaOH is approximately 3.58.

Step-by-step explanation:

To determine the volume of base required to reach the equivalence point in the titration of HF with NaOH, we need to consider the stoichiometry of the reaction: HF + NaOH → NaF + H2O. The balanced equation tells us that 1 mole of HF reacts with 1 mole of NaOH. Therefore, the volume of NaOH required to reach the equivalence point is equal to the volume of HF used, which is 50 mL. So the answer is (b) 50 mL.

To calculate the initial pH of the HF solution, we can use the Ka expression for HF: Ka = [H+][F-]/[HF]. Since the initial concentration of HF is 0.180 M and we're assuming it's a weak acid dissociating to a small extent, we can neglect the change in concentration of HF and assume that [HF] remains constant.

Therefore, we can rewrite the Ka expression as Ka = [H+][F-]/(0.180). Rearranging the equation gives us [H+] = Ka * [HF]/[F-]. Substituting the given values, we have [H+] = 3.5 x 10^-4 * 0.180/0.180 = 3.5 x 10^-4. Taking the negative logarithm gives us the pH of 3.46. So the initial pH of the HF solution is approximately 3.46.

To calculate the pH after adding 5.0 mL of NaOH, we need to consider that NaOH is a strong base and will react completely with HF to form NaF and water. Since 1 mole of NaOH reacts with 1 mole of HF, we can calculate the concentration of NaOH that has reacted.

The initial concentration of NaOH is 0.210 M, and we've added 5.0 mL, so the number of moles of NaOH that has reacted is 0.210 M * 0.005 L = 0.00105 mol. Since this is the same as the number of moles of HF that has reacted, the concentration of HF remaining is 0.180 M - 0.00105 mol/0.050 L = 0.157 M.

Using the same Ka expression as before, we can calculate the concentration of H+ ions in the solution, which is equal to the pH. [H+] = Ka * [HF]/[F-] = 3.5 x 10^-4 * 0.157/0.210 = 2.620 x 10^-4. Taking the negative logarithm gives us the pH of 3.58. So the pH after adding 5.0 mL of NaOH is approximately 3.58.