Question

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?

Express your answer with the appropriate units

Answer 1

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 %