How many moles of NaF should be added to 1.75 L of 0.0392 mol L−1 HF(aq) to produce a solution with a pH of 3.70 ? Assume no change in volume and a temperature of 25 °C.For HF, pKa = 3.18 at 25 °C. Enter a number accurate to 3 significant figures. To produce a solution of the desired pH:    Add   moles of NaF.

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

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

where:

• pH is the pH of the solution
• pKa is the acid dissociation constant
• [A-] is the concentration of the base (F- in this case)
• [HA] is the concentration of the acid (HF in this case)

Given:

• pH = 3.70
• pKa = 3.18
• [HA] = 0.0392 mol/L
• Volume of solution = 1.75 L

We need to find [A-], the concentration of F-.

Rearranging the Henderson-Hasselbalch equation to solve for [A-], we get:

[A-] = [HA] * 10^(pH - pKa)

Substituting the given values:

[A-] = 0.0392 mol/L * 10^(3.70 - 3.18)

[A-] = 0.0392 mol/L * 10^0.52

[A-] = 0.0392 mol/L * 3.311

[A-] = 0.1298 mol/L

Since we assumed no change in volume, the moles of NaF added is equal to the moles of F- in the solution. Therefore, the moles of NaF added is:

Moles of NaF = [A-] * Volume

Moles of NaF = 0.1298 mol/L * 1.75 L

Moles of NaF = 0.227 moles

Therefore, to produce a solution with a pH of 3.70, you should add approximately 0.227 moles of NaF.