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What is the pH of a 1.0 M solution of potassium formate, KHCO2? (No equilibrium constant...

What is the pH of a 1.0 M solution of potassium formate, KHCO₂? (No equilibrium constant given).

Expert Solution

HCO2^- + H2O ------------------> HCO2H + OH^-

I 1 0 0

C -x +x +x

E 1-x +x +x

Kb = Kw/Ka

= 1*10^-14/1.8*10^-4

= 5.6*10^-11

Kb = [HCO2H][OH-]/[HCO2^-]

5.6*10^-11 = x*x/(1-x)

5.6*10^-6*(1-x) = x^2

x = 0.0024

[OH-] = x = 0.0024M

POH = -log[OH-]

= -log0.0024

= 2.6197

PH = 14-POH

= 14-2.6197

= 11.3803 >>>>>answer

queen_honey_blossom answered 1 year ago

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