Question

A metal hydroxide of molecular formula $M(OH{)}_4$ is $50$% ionised. Its $0.0025M$ solution will have the $pH$.

• A. $12$
• B. $2$
• C. $4$
• D. $11.7$

Answer 1

Answer: (D)

$11.7$

Ionization of metal hydroxide can be represented as:

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

initial concentrations:

$0.0025M0M0M$

at equillibrium:

$(0.0025-x)MxM4xM$

Given that 50% ionized,

$\therefore x=\frac{50}{100}\times 0.0025=1.25\times {10}^{-3}M$

Thus $\left[O{H}^{-}\right]=4x=4\times 1.25\times {10}^{-3}=5\times {10}^{-3}M$

$pOH$=-log([$O{H}^{-}$])

$pOH$=-log ($5\times {10}^{-3}$)

$pOH$=2.3

Using $pOH+pH=14$

we get,

$pH=14-2.3=11.7$

Thus pH of the solution is 11.7.