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

Answer:

a) The pH of the solution is 12.13.

b) The pH of the solution is 12.17.

Explanation:

Ionic product of water =$K_w=1.01\times 10^-{14}$

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

$1.01\times 10^-{14}=[H^+][OH^-]$

Taking negative logarithm on both sides:

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

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) $1.39\times 10^{-2} M$ of NaOH.

Concentration of hydroxide ions:

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

So, $[OH^-]=1\times [NaOH]=1\times 1.39\times 10^{-2} M=1.39\times 10^{-2} M$

$pOH=-\log[1.39\times 10^{-2} M]=1.86$

$13.99=pH+pOH$

$13.99=pH+1.86$

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, $[OH^-]=3\times [Al(OH)_3]=3\times 0.0051 M=0.0153 M$

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

$13.99=pH+pOH$

$13.99=pH+1.82$

pH=13.99-1.82=12.17