Question

You add 30ml of 1M of NH3 to a 50ml Solution of 0,2 MgCl2. Magnesium hydroxide (Mg(OH)2) precipitates.

What is the minimum concentration of NH4+ that prevents the precipitation of Mg(OH)2?

Given information:
(MgOH2) Ksp = 13
(NH3) Kb = 10^-4.75



Note: a complex formation between Mg and NH4+ is ruled out.

Answer 1

Step-by-step explanation:

The reaction equation will be as follows.

`NH_(3).H_(2)O(aq) + MgCl_(2)(aq) \rightleftharpoons Mg(OH)_(2) + NH_(4)Cl`

`K_(sp)` of
`Mg(OH)_(2)` = 13 and
`K_(b)` of
`NH_(3)` =
`10^(-4.75)`

Hence, concentration of
`MgCl_(2)` =
`\frac{\text{no. of moles}}{\text{volume in liter}}`

=
`(0.2 mol)/(50 * 10^(-3))`

= 4 M

As,
`Mg(OH)_(2) \rightleftharpoons Mg^(2+) + 2[OH^(-)]`

Hence,
`K_(sp) = [Mg^(2+)][OH^(-)]^(2)`

13 =
`4 * [OH^(-)]^(2)`

`[OH^(-)]` = 1.80276 M

Also, ionization constant of water is given as follows.

`K_(w) = [H_(3)O^(+)][OH^(-)]` ......... (1)

`K_(w) = 10^(-14)`

Now, put the value of
`[OH^(-)]` in equation (1) as follows.

`K_(w) = [H_(3)O^(+)][OH^(-)]`

`[OH^(-)] = (K_(w))/([H_(3)O^(+)])`

=
`(10^(-14))/(1.80276)`

=
`0.554 * 10^(-14) M`

Also,
`K_(w) = K_(a) * K_(b)` =
`10^(-14)`

Therefore,
`K_(a)` =
`(10^(-14))/(k_(b))`

=
`(10^(-14))/(10^(-4.75))`

=
`10^(-9.75)`

Since,
`H_(2)O(l) + NH^(+)_(4) \overset{K_(a)}{\rightleftharpoons} H_(3)O^(+)(aq) + NH_(3)(aq)`

`K_(a) = ([H_(3)O^(+)[NH_(3)]])/([NH^(+)_(4)])` ....... (2)

Hence, calculate the value of
`[NH^(+)_(4)]` as follows.

`[NH^(+)_(4)]` =
`([H_(3)O^(+)][NH_(3)])/(K_(a))`

=
`(0.554 * 10^(-14) * 1)/(10^(-9.75))`

=
`0.554 * 10^(-4.75) M`

`[NH^(+)_(4)]` =
`3.115 * 10^(-5) M`

Thus, we can conclude that the minimum concentration of
`[NH^(+)_(4)]` that prevents the precipitation of
`Mg(OH)_(2)` is
`3.115 * 10^(-5) M`.