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# The pH of a 0.050 mol/L solution of ammonia is ?

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- pisgahchemistLv 76 months agoFavorite Answer
pH of ammonia solution ....

NH3(aq) + HOH(l) <==> NH4^+ + OH- ........... Kb = 1.8x10^-5

0.050M ............................ 0 .......... 0 .............. initial

-x ................................... +x ......... +x ............. change

0.050-x ............................ x ........... x ............. equilibrium

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

1.8x10^-5 = x² / (0.050-x)

x = 9.4x10^-4

[OH-] = 9.4x10^-4M

pH = -log[H+] = -log(Kw / [OH-]) = -log(1.00x10^-14 / 9.4x10^-4) = 10.97

The pH is expressed to two significant digits. When taking the log of a number, only the digits to the right of the decimal reflect the precision in the original number.

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The answer is accurate. Don't be put off by the "thumbs down." It's the work of a troll.

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