Question

# THIS IS MY SECOND TIME POSTING THIS QUESTION. PLS MAKE SURE ANSWERS ARE CORRET Ammonia, NH3,...

THIS IS MY SECOND TIME POSTING THIS QUESTION. PLS MAKE SURE ANSWERS ARE CORRET

Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5.

Part A What is the pH of a 0.490 M ammonia solution? Express your answer numerically to two decimal places.

Part B

What is the percent ionization of ammonia at this concentration?

A)

the pH of [NH3] = 0.49 M should be calculated as follows:

NH3 + H2O <-> NH4++ OH-

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

initially

[NH4+] = 0

[OH-] = 0

[NH3] = 0.49

in equilibrium

[NH4+] = +x

[OH-] = +x

[NH3] = 0.49 - x

substitute in Kb

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

1.8*10^-5 = (x*x)/(0.49-x)

solve for x

x =[OH-] = 0.00296

pOH = -log(OH) = log(0.00296)

pOH = 2.5287

pH = 14-pOH = 14-2.5287

pH = 11.4713 --> 2 decimals --> 11.47

B)

Find % ionization of ammonia

% ioniazation = [NH4+] / [NH3]0 * 100%

substitute data

% ioniazation = 0.00296/0.49 * 100% = 0.6040816 % --> 0.6 %

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