Question

You desire to create a solution with a pH of 3.03. If you add 0.533 moles of HF to 1.00 L of solution, how many moles of NaF should you add?

Answer 1

PH = -log [H+]

[H+] = 10^-PH

Accordind to the question the PH is 3.03

[H+] = 10^-3.

According to the question

MV =M1V1-M2V2 M1V1 is for the HF

M2V2 is for the NaF

MV is fo the resultant solution.

Here the volume is constant, That means it is the one litre that is given in the question.

Then consider the concentrations only. (here the volume is one litre then consider as moles).

MV =M1V1-M2V2 Molarity = moles x 1000/ Volume in mL (or)

Molarity = moles/litre.

C = C1 - C2

10^-3 = 0.533- C₂

0.001 = 0.533-C₂

0.001-0.533 = -C₂

-C₂ = -0.532

C₂ = 0.532 moles.