Question

In: Chemistry

What is the pOH of 0.0042 M LiF? Ka = 7.1 x 10-4 7.39 11.24 6.61...

What is the pOH of 0.0042 M LiF?

Ka = 7.1 x 10-4

7.39

11.24

6.61

2.76

9.15

Expert Solution

LiF -----------> F- + Li+

F^- + H2O------------> HF + OH^-

I 0.0042 0 0

C -x +x +x

E 0.0042-x +x +x

Kb = Kw/Ka = 1*10^-14/7.1*10^-4 = 1.4*10^-11

Kb = [HF][OH-]/[F-]

1.4*10^-11 = x*x/0.0042-x

1.4*10^-11*(0.0042-x) = x^2

x = 2.42*10^-7

[OH-] = x = 2.42*10^-7M

POH = -log[OH-]

= -log2.42*10^-7 = 6.6161 >>>>>>answer

queen_honey_blossom answered 2 years ago

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