Question

Determine the pH of the following base solutions. (Assume that all solutions are at 25°C and the ion-product constant of water, Kw, is 1.01 ✕ 10−14.) (a) 1.39 ✕ 10−2 M NaOH WebAssign will check your answer for the correct number of significant figures. (b) 0.0051 M Al(OH)3 WebAssign will check your answer for the correct number of significant figures.

Answer 1

a) The pH of the solution is 12.13.

b) The pH of the solution is 12.17.

Step-by-step explanation:

Ionic product of water =
`K_w=1.01* 10^-{14}`

`K_w=[H^+][OH^-]`

`1.01* 10^-{14}=[H^+][OH^-]`

Taking negative logarithm on both sides:

`-\log[1.01* 10^-{14}]=(-\log [H^+])+(-\log [OH^-])`

The pH is the negative logarithm of hydrogen ion concentration in solution.

The pOH is the negative logarithm of hydroxide ion concentration in solution.

`13.99=pH+pOH`

a)
`1.39* 10^(-2) M` of NaOH.

Concentration of hydroxide ions:

`NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)`

So,
`[OH^-]=1* [NaOH]=1* 1.39* 10^(-2) M=1.39* 10^(-2) M`

`pOH=-\log[1.39* 10^(-2) M]=1.86`

`13.99=pH+pOH`

`13.99=pH+1.86`

pH=13.99-1.86=12.13

b)
`0.0051 M` of NaOH.

Concentration of hydroxide ions:

`Al(OH)_3(aq)\rightarrow Al^(3+)(aq)+3OH^-(aq)`

So,
`[OH^-]=3* [Al(OH)_3]=3* 0.0051 M=0.0153 M`

`pOH=-\log[0.0153 M]=1.82`

`13.99=pH+pOH`

`13.99=pH+1.82`

pH=13.99-1.82=12.17