In: Chemistry
Part A
How many moles of HF (Ka=6.8×10−4) must be added to water to form 0.230 L of solution with a pH of 2.50?
Express your answer using two significant figures.
n=____mol
Answer – We are given, pH = 2.50 , volume = 0.230 L , Ka = 6.8*10-4
We know,
[H3O+] = 10-pH
= 10-2.50
= 0.00316 M
We need to put ICE chart and assume [HF] = Y
HF + H2O <-----> H3O+ + F-
I Y 0 0
C -x +x +x
E Y-x 0.00316 +x
So at equilibrium, x = [H3O+] = 0.00316 M
So,
Ka = [H3O+][F-] / [HF]
6.8*10-4 = 0.00316 *0.00316 / (Y-0.00316)
6.8*10-4 (Y-0.00316) = 1.0*10-5
6.8*10-4 Y -2.15*10-6 = 1.0*10-5
6.8*10-4 Y = 1.0*10-5 + 2.15*10-6
= 1.21*10-5
So, Y = 0.0179 M
So, [HF] = 0.0179 M
We know
Moles of HF = 0.0179 M * 0.230 L
= 0.00411 moles of HF
0.00411 moles of HF (Ka=6.8×10−4) must be added to water to form 0.230 L of solution with a pH of 2.50