Question

Calcium hydroxide is a strong base. Compute $\left[C{a}^{2+}\right]$ and $\left[O{H}^{-}\right]$ for a solution that is prepared by dissolving $0.60g$ of $Ca(OH{)}_2$ in enough water to make a $1500$ mL of solution.
[Atomic weights: $Ca=40,O=16,H=1$]

• A. $5.4\times {10}^{-3},9.1\times {10}^{-13}$
• B. $5.4\times {10}^{-3},1.08\times {10}^{-2}$
• C. $5.4\times {10}^{-3},5.4\times {10}^{-3}$
• D. $8.1\times {10}^{-3},8.1\times {10}^{-3}$

Answer 1

Answer: (B)

$5.4\times {10}^{-3},1.08\times {10}^{-2}$

No. of moles of $Ca(OH{)}_2$ is $0.6/74$.

So, molarity is $\frac{0.6}{74}\times \frac{1000}{1500}=5.4\times {10}^{-3}$

As each mole produces one mole of $\left[C{a}^{2+}\right]$ and two moles of $O{H}^{-}$ ion.

$\left[C{a}^{2+}\right]=5.4\times {10}^{-3}$

$\left[O{H}^{-}\right]=2\times 5.4\times {10}^{-3}=1.08\times {10}^{-2}$