Question

33. Calculate the pH of a solution that is 1.00 M HF, 2.00 M HCl, and...

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

Homework Answers

Answer #1

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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