Question

Part A

Calculate the pH of 0.100 L of a buffer solution that is 0.23 M in HF and 0.55 M in NaF.

Part B

Calculate pH of the solution on addition of the following. 0.004 mol of HNO3

Part C

Calculate pH of the solution on addition of the following. 0.002 mol of KOH

Answer 1

Part A

They have given pH , volume of HF, and NaF. We can use Henderson Hasselbalch equation and find out the pH.

Equation

Lets plug given values in above equation and get the value of pH.

Pka = - log ka

Ka of HF = 6.6E-04

Therefore pka = -log ( 6.6 E-4 )

= 3.18

Lets use this value

Therefore the pH of the solution would be 3.55

Part B

In this part we have to find pH after addition of 0.004 moles of HNO3

Since HNO3 is acid so it will reacts with base and forms acid (HF) back.

Lets show the reaction of HNO3 with base.

Number of moles HF formed equal to number of moles of F- disappears and both are equal to moles of HNO3.

Now lets calculated equilibrium molarity of HF and F

[NaF-] = [F-] = 0.51 M

Lets use Same equation (Henderson Hasselbalch euqution)

Part C

Now we have to use moles of base and find pH. We use same procedure. Here moles of acid will decrease and moles of base will decrease. Lets show the reaction.

KOH + HF--- > KF (aq) + H2O (l)

This gives mole ration (1:1)

So the pH of this solution is 3.61