What is the pH of a 0.445 M ammonia solution? Express your answer numerically to two...

What is the pH of a 0.445 M ammonia solution? Express your answer numerically to two decimal places.

Expert Solution

NH3(aq) + H2O(l) -------------> NH4^+ (aq) + OH^- (aq)

I 0.445 0 0

C -x +x +x

E 0.445-x +x +x

Kb = [NH4^+][OH^-]/[NH3]

1.8*10^-5 = x*x/(0.445-x)

1.8*10^-5 *(0.445-x) = x^2

x = 0.00282

[OH^-] = x = 0.00282M

POH = -log[OH^-]

= -log0.00282

= 2.5497

PH = 14-POH

= 14-2.5497

= 11.45 >>>>>answer

queen_honey_blossom answered 1 year ago

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