Question

Calcium hydroxide is a strong base. Compute ${{[Ca]}^{2+}}$and${{[OH]}^{-}}$ for a solution that is prepared by dissolving 0.60g of $Ca{{(OH)}_{2}}$ in enough water to make a 1500mL 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

Hint: The answer for this question includes the calculation of concentrations of the calcium and hydroxide ions by firstly calculating the molecular weight of $Ca{{(OH)}_{2}}$ and followed by using the formula of molarity that is given by $Molarity=\dfrac{{{n}_{solute}}}{{{V}_{solution(Litres)}}}$

Complete answer:
We have studied in our chapters of chemistry in the lower classes that deal with the basic concepts and the terminologies that are molarity, molality and also normality of the solutions.
Now, let us see in detail about this which helps us to deduce the required answer.
Firstly, let us calculate the molecular weight of the compound given that is $Ca{{(OH)}_{2}}$
$Mol.wt.\left[ Ca{{(OH)}_{2}} \right]=40+2\left( 16 \right)+2(1)=74g/mol$
Now, we have the data given as the weight of calcium hydroxide that is 0.60g that is w=0.60 g
The volume of the solution in litres is V = 1500 ml = 1.5 L
Now, the molarity of the solution according to the formula will be,
$Molarity=\dfrac{{{n}_{solute}}}{{{V}_{solution(Litres)}}}$
Here, number of moles of solute that is calcium hydroxide is,${{n}_{solute}}=\dfrac{w}{Mol.wt.}=\dfrac{0.6}{74}=0.008moles$
Substituting the values in above equation, we get
\[Molarity=\dfrac{0.008}{1.5}=5.4\times {{10}^{-3}}M\]
Now, dissociation of calcium hydroxide will be,
\[Ca{{(OH)}_{2}}\to C{{a}^{2+}}+2O{{H}^{-}}\]

Initial conc.	\[5.4\times {{10}^{-3}}\]	0	0
Final conc.	0	\[5.4\times {{10}^{-3}}\]	\[2\times 5.4\times {{10}^{-3}}\]

Now, according to the above calculation we have,
\[[C{{a}^{2+}}]=5.4\times {{10}^{-3}}M\]
\[{{[OH]}^{-}}=2\times 5.4\times {{10}^{-3}}=1.08\times {{10}^{-2}}M\]

Thus, the correct answer is option B).

Note: Note that while solving the problems that are based on the molality and molarity of the solutions, do not be confused about the two terms as molality is the measure of number of moles of solute present in 1 kg of the solvent and molarity is the measure of number of moles of solute present in 1 litre of the solution.