# Question #01170

##### 1 Answer

#### Answer:

The

#### Explanation:

Because ammonium

**Step 1**: find

#K_"a" * K_"b" = 10^-14#

#K_"a"# of#("NH"_4)^(+)=10^-14/(K_text(b))=5.6 xx 10^-10#

**Step 2**: Set up the dissociation of ammonium equilibrium, and find the amount of

#{: (," ",bb("NH"_4^+),bb->,bb"NH"_3,+,bb("H"^+)), ("initial:",,"0.100 M",,"0 M",,"0 M"), ("change:",,-x,,+x,,+x), ("equilibrium:",,0.1-x,,x,,x) :}#

You can approximate that

#0.1 - x ~~ 0.1#

b/c

#K_"a" = (["NH"_3] * ["H"^+])/(["NH"_4^+]) = (x * x) /0.1 = 5.6 xx 10^-10#

#x^2 = 0.1 * 5.6 xx 10^-10#

#x = sqrt(5.6 xx 10^-11) =7.5 xx 10^-6#

#["H"^+] = x = 7.5 xx 10^-6#

**Step 3**: calculate

#"pH" = -log["H"^+] = -log(7.5xx10^-6) = 5.1#