Question

What is the pH of $0.0067M$$KOH$ solution?

Answer 1

Hint: $pH$ is a measure used in chemistry to describe the acidity or basicity of an aqueous solution. Acidic solutions (those containing a higher concentration of ${H^ + }$ ions) have a lower $pH$ than basic or alkaline solutions.

Complete answer:
$KOH$ is the formula name of Potassium hydroxide. It is colorless liquid and acts like a strong base (alkali). When $KOH$ dissolves with water, it completely dissociates into its ion. It breaks into its constituent anion and cation - ${K^ + }$ and $O{H^ - }$. The charges on these ions are then more stabilized by being fully solvated (i.e. surrounded by water molecules).
The reaction of potassium hydroxide with water and the dissociation of its constituent ions is represented by this equation –
$KOH \to {K^ + } + O{H^ - }$
Firstly, we will find the relationship between $KOH$ and $O{H^ - }$ ion. It is divided in the equal $1:1$ ratio.
$1:1$ ratio $ \Rightarrow [KOH] = [O{H^ - }] = 6.7 \times {10^{ - 3}}M$
$[{H_3}{O^ + }]$ and $[O{H^ - }]$ must follow the given condition in the aqueous solution –
\[[{H_3}{O^ + }][O{H^ - }] = {K_w}\]
Now shifting $[O{H^ - }]$ to the denominator of the right-hand side, then the equation is as follows –
$[{H_3}{O^ + }] = \dfrac{{{K_W}}}{{[O{H^ - }]}}$
Now, by this we can find the molar concentration of hydronium ion.
Since we know the values of molar concentration of ${K_w}$ and $O{H^ - }$ ion.
Molar concentration of ${K_w}$ is $1 \times {10^{ - 14}}$.
Molar concentration of $[O{H^ - }]$ ion is $6.7 \times {10^{ - 3}}M$.
Substituting the values in the equation, we get –
$[{H_3}{O^ + }] = \dfrac{{1 \times {{10}^{ - 14}}}}{{6.7 \times {{10}^{ - 3}}}}$
$[{H_3}O] = 1.5 \times {10^{ - 12}}M$
Since we got to know the concentration of hydronium ion, now we can determine the pH of the $0.0067M$$KOH$ solution –
pH $ = - \log [{H_3}{O^ + }]$
pH = $ - \log [1.5 \times {10^{ - 12}}]$
Checking the value of pH from the pH table, we get –
pH $ = 11.83$
Hence, pH value of $0.0067M$ $KOH$ solution is $11.83$.

Note:
At $25^\circ C$, acidic solutions have a pH less than $7$, whereas basic solutions have a pH greater than $7$. At this temperature, pH$ = 7$ solutions are neutral (e.g. pure water). The pH neutral value changes with temperature, being less than $7$ as the temperature rises.