Question

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?

Answer 1

`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