Question

How many moles of $Ca{\left(OH\right)}_2$ must be dissolved to produce $250mL$ of an aqueous solution of $pH$ $10.65$ assuming complete dissociation?

• A. $5.6\times {10}^{-5}$ $mol$
• B. $6.5\times {10}^{-5}$ $mol$
• C. $6.9\times {10}^{-5}$ $mol$
• D. None of these

Answer 1

The correct option is A $5.6\times {10}^{-5}$ $mol$

SInce dissociation is complete, hence, $\alpha =1$
Now, we know that,
$pH+pOH=14$
$pOH=14-pH=14-10.65=3.35$
$pOH=-lo{g}_{10}\left[O{H}^{-}\right]$
$\left[O{H}^{-}\right]=Antilog(-3.35)$ or ${10}^{-3.35}=4.46\times {10}^{-4}mol/d{m}^3$
Now,
$\left[O{H}^{-}\right]=n\alpha C$
where, n=Acidity of the given base or no. of replaceable $O{H}^{-}$ ion=2
Hence,
$C=\frac{4.46\times {10}^{-4}}{2\times 1}=2.23\times {10}^{-4}$
We know that
C= Molarity of the solution$=\frac{\text{No. of moles of solute}}{{\text{Volume of solution in dm}}^3}$
Hence, no. of moles of $Ca(OH{)}_2$=$2.23\times {10}^{-4}\times 0.25=5.57\times {10}^{-5}mol$