Question

A metal hydroxide of the molecular formula \[M{\left( {OH} \right)_4}\] ​ is $50\% $ ionized. Its \[0.0025\;\] M solution will have the pH.
A. $12$
B. $2$
C. $4$
D. $11.7$

Answer 1

Hint: pH of a solution is the negative of the base $10$ logarithm of hydronium ion concentration in a solution. The concentration of hydronium ions in an aqueous solution can provide information about the acidic, basic, or neutral nature of a solution. For an acidic solution, pH is less than $7$ , and for a basic solution, pH is greater than $7$. At room temperature, the pH of neutral water is equal to $7$

Complete step by step answer:

Ionization of metal hydroxide can be represented as:

$M{(OH)_4} \to {M^ + } + 4O{H^ - }$

	$M{(OH)_4}$	${M^ + }$	$4O{H^ - }$
INITIAL CONC.	$0.0025M$	$0M$	$0M$
CONC AT EQUILIBRIUM	$(0.0025 - x)M$	$xM$	$4xM$

Given that $50\% $is ionized,
$x = \dfrac{{50}}{{100}} \times 0.0025 = 1.25 \times {10^{ - 3}}M$
Thus $\left[ {O{H^ - }} \right] = 4 \times 1.25 \times {10^{ - 3}}M$
$ = 5 \times {10^{ - 3}}M$
To calculate the pH of an aqueous solution, you need to know the concentration of hydronium ions in moles per litre (molarity) of the solution.
pH is then calculated using this expression
$pH = - {\log _{10}}\left[ {{H_3}{O^ + }} \right]$
Likewise, $pOH = - {\log _{10}}(O{H^ - })$
$ \Rightarrow $$pOH = - {\log _{10}}(O{H^ - })$
$ = - {\log _{10}}(5 \times {10^{ - 3}})$
$ \Rightarrow $$pOH = 2.3$

We know,
$ \Rightarrow $$pH + pOH = 14$
$pH = 14 - 2.3$
$ = 11.7$

So the correct answer is D.

Note: Molarity is defined as the number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.
Molarity$ = $No. of moles of solute/volume of solution in litre
To determine the strength of acids and bases, we use a universal indicator i.e. pH which shows a different concentration of hydronium ion concentration in a solution.
The pH scale ranges from $0 - 14$.
Glass electrode, pH metre, or colour changing indicator may be used to measure the pH of an aqueous solution