Question

How many moles of KF should be added to 0.50 L of a 0.10 M HF solution to prepare a buffer with a pH of 2.84?

Answer 1

Given

HF      

Molarity= 0.1

volume            =0.5 L

pH buffer = 2.84

Solution:

Lets write reaction for this buffer

            HF + KOH -- > KF + H2O

            We first calculate concentration of F^- by using Henderson Hasselbalch equation.

The equation :

pH = pka + log([conj. Base]/[weak acid])

here conjugate base is F^- and weak acid is HF

pka = -log ka

ka of HF = 7.2 E-4

pka = -log (7.2E-4)

pka = 3.143

lets plug all the values in Henderson equation

2.84 = 3.143 + log ([F^-] /0.1)

log ([F^-] /0.1) = 2.84 – 3.143

log ([F^-] /0.1) =-0.3026675

Lets take antilog of both side and find the value of [F-]

[F-]= 0.0498117 M

We know [F-] = [KF-]   ; KF dissociates fully.

We use molarity and volume in L to calculate the moles of F-

Mol = Molarity x volume in L

n KF= 0.0498117 M x 0.5 L

            =0.025 mol

So we need to add 0.025 moles of KF to the solution.