33. Calculate the pH of a solution that is 1.00 M HF, 2.00 M HCl, and 1.439 MNaF. (Ka = 7.2 10–4)
3.14
3.30
2.48
2.98
0.25
34. If the solid CuF2 has a solubility of 0.0020 mol/L, what is the value of Ksp?
1.8 x 10-7
4.0 x 10-6
3.2 x 10-8
8.0 x 10-9
none of the above
33)
Molarity of HF = 1 M
Molarity of HCl = 2 M
Molarity of NaF = 1.439 M
Ka = 7.2 x 10-4
[F-] = 1.439 -1 = 0.439 M
We have,
pH = -logKa + log[F-]/[HCl] = -log(7.2 x 10-4) + log[0.439]/[2]
= 3.143 – 0.659 = 2.48
34)
CuF2 Cu2+ + 2F-
x x 2x
Ksp = [Cu2+][F-]2 = 4x3 = 4 x (0.0020)3 = 3.2 x 10-8
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