Question

What is the $pH$ of a $1.4\cdot {10}^{-2}MNaOH$ solution?

Answer 1

$pH=12.15$
Explanation:
Even before doing any calculations, you can say that since you're dealing with a strong base, the $pH$ of the solution must be higher than 7.
The higher the concentration of the base, the higher the pH will be.
In your case, you're dealing with a solution of sodium hydroxide, $NaOH$, a strong base that dissociates completely in aqueous solution to form sodium cations, $N{a}^{+}$, and hydroxide anions, $O{H}^{-}$
$NaO{H}_{\left(aq\right]}\to N{a}_{\left(aq\right]}^{+}+O{H}_{\left(aq\right]}^{-}$
Notice that the salt dissociates in a 1:1 mole ratio with the hydroxide anions, you can say that
$\left[O{H}^{-}\right]=\left[NaOH\right]=1.4\cdot {10}^{-2}M$
Now, the $pH$ of the solution is determined by the concentration of hydronium ions, $H_3{O}^{+}$. For aqueous solutions, the concentration of hydronium ions is related to the concentration of hydroxide ions by the ion product constant of water, $K_WK_W=[OH^−]\cdot [H_3O^+]$Atroomtemperature,youhave$K_W=10^{−14}$
This means that the concentration of hydronium ions can be determined by using
${[}_{H}3O^{+}]=\frac{K_W}{\left[O{H}^{-}\right]}$
Plug in your values to get
$\left[H_3{O}^{+}\right]$=$\frac{{10}^{-14}}{1.4\times {10}^{--2}}$=$7.14\cdot {10}^{-13}M$
The $pH$ of the solution is equal to
$pH=-\log \left(\right[H_3{O}^{+}\left]\right)$
In your case,
$pH=-\log (7.14\cdot {10}^{-13})=12.15$
As predicted, the pH is not only higher than 7, but it is significantly higher than 7.
Alternatively, you can use the $pOH$ of the solution to find its pH. As you know,
$pOH=-\log \left(\right[O{H}^{-}\left]\right)$
In your case,
$pOH=-\log (1.4\cdot {10}^{-2})=1.85$
You know that
$pH+pOH=14$
and so, once again, you have
$pH=14-1.85=12.15$