Question

For a solution that is 0.162 M NH3 and 0.102 M NH4Cl calculate the following.

Part A

[OH−]

Express your answer using two significant figures.

Part B

[NH+4]

Express your answer using three significant figures.

Part C

[Cl−]

Express your answer using three significant figures.

Part D

[H3O+]

Express your answer using two significant figures.

Answer 1

POH   = Pkb + log[NH4Cl]/[NH3]

          = 4.74 + log0.102/0.162

        = 4.74-0.2009 = 4.5391

POH = 4.5391

-log[OH-] = 4.5391

     [OH-] = 10^-4.5391   = 2.9*10^-5 M

part B

NH4Cl --------> NH4⁺ +    Cl^-

0.102M             0.102M        0.102M

[NH4⁺] = 0.102M

part-C

[Cl-]   = 0.102M

part-D

[H3O+]   = Kw/[OH-]

              = 1*10^-14/2.9*10^-5

              = 3.4*10^-10 M