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

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

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4.1 Ratings (677 Votes)

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

IÂ Â Â Â Â Â Â Â Â Â Â Â Â 0.445Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

CÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +x

EÂ Â Â Â Â Â Â Â Â Â 0.445-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +x

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â KbÂ Â Â =Â Â Â [NH4^+][OH^-]/[NH3]

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

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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â = 0.00282

Â Â Â Â Â Â Â Â Â Â [OH^-]Â Â Â = xÂ Â = 0.00282M

Â Â Â Â Â Â Â Â Â POHÂ Â Â Â = -log[OH^-]

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = -log0.00282

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2.5497

Â Â Â Â Â Â Â Â PHÂ Â Â Â Â Â Â Â Â = 14-POH

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 14-2.5497

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 11.45 >>>>>answer

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

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