Calculate the pH of .100 of a buffer solution that is .25 M in HF and .50 M in NaF. What is the change in pH on addition of the following?

Question

A. .002 mol of HNO3
B. >004 mol of KOH

I calculated the correct pH of the solution but am having trouble calculating A & B.

Thanks,
Ivy

Answer 1

What's that 0.100? 0.100 L?
millimols HF = mL x M = 100 x 0.25 = 25.0
millimols NaF = 100 x 0.500 = 50,0
Add 0.002 = 2 millomils HNO3

.........F^- + H^+ ==> HF
I.......50.0...0......25.0
add.............2..........
C........-2....-2......+2
E........48.....0.....27

Then plug the E line values into the HH equation and solve for pH.

Addition of NaOH is done the same way but use the reverse chemical equation since this is a base added.

.........HF + OH^- ==> F^- + H2O