In: Chemistry
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?
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.