Question

In: Chemistry

determine the [OH-] and pH of a solution tht is 0.140 M in F-.

Expert Solution

Answer –

We are given, [F^-] = 0.140 M

First we need to look the Kb value for F^- from the Ka value of HF

The Ka value for HF = 7.2*10^-4

So, Kb for F^- = 1*10^-14 / Ka

= 1*10^-14 / 7.2*10^-4

= 1.39*10^-11

Now we need to put ICE chart

F^- + H₂O -----> HF + OH^-

I 0.140 0 0

C -x +x +x

C 0.140-x +x +x

We know,

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

1.39*10^-11 = x*x/(0.140-x)

We can neglect x in the 0.140-x, because Kb value is too small

^x2 = 1.39*10^-11 *0.140

= 1.92*10^-12

x = 1.39*10^-6 M

so, x = [OH^-] = 1.39*10^-6 M

now we can calculate pOH from [OH^-]

pOH = -log [OH^-]

= - log 1.39*10^-6 M

= 5.86

So, pH = 14- pOH

= 14 -5.86

= 8.14

queen_honey_blossom answered 19 hours ago

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