If #"0.50 M"# of #"NH"_4"OH"# were to dissociate in water, and #K_b = 1.773 xx 10^(-5)#, what is the resultant #"pH"# at equilibrium?

1 Answer

Answer:

#pH=11.47#

Explanation:

Let's make an (R)ICE table for the dissociation of #NH_4OH# (the base you get when you dissolve #NH_3# in water.)

#NH_3+H_2O rightleftharpoons NH_4OH#

Suppose that #x# #M# of #NH_4OH# then dissociated. (#x>=0#)

Reaction: #NH_4OH " "rightleftharpoons" " NH_4^+ + OH^-#

Initial: #" "" ""0.50 M"##" "" "" "##"0 M"##" "" "" "##"0 M"#

Change: #" "-x" M"##" "##" "+x# #"M"" "" "# #+x# #"M"#

Equil.: #" "(0.50-x)# #"M"##" "##x# #"M"" "" "# #" "x# #"M"#

Using the mass action expression for #K_b#:

#K_b=([NH_4^(+)][OH^-])/([NH_4OH]) = (x^2)/(0.50-x)#

Solve for #x#:

given that #K_b=1.773*10^-5#,

#x=0.002969# #M#

Thus #[OH^-]=0.002969# #M#

Evaluate the #pH#:

#color(blue)(pH) = 14 - pOH = 14 + log[OH^-] = color(blue)(11.47)#

to two sig figs. :)

Source: https://www.ck12.org/chemistry/Calculating-Ka-and-Kb/lesson/Calculating-Ka-and-Kb-CHEM/?referrer=featured_content