Question

50gm of sample of $Ca{{(OH)}_{2}}$ is dissolved in 50 ml of 0.5N HCl solution. The excess HCl was titrated with 0.3N NaOH. The volume of NaOH was 20cc. Calculate the % purity of $Ca{{(OH)}_{2}}$.

Answer 1

Hint: Firstly, find out the moles of HCl reacted and then use it to find the moles of calcium hydroxide. Calculate the mass of calcium hydroxide from the number of moles by multiplying it by molar mass. Then you can find out the percentage purity of calcium hydroxide by using the formula-
     \[Percentage\text{ purity=}\dfrac{Mass\text{ obtained}}{Mass\text{ of sample}}\times 100\]

Complete step by step answer:
 In the question it is given to us that a sample of calcium hydroxide is added in a 50ml 0.5N HCl solution.
Firstly, let us calculate the number of moles of sodium hydroxide used. Here, the concentration of the NaOH solution is given to us in terms of normal but as NaOH has only 1 hydroxyl ion therefore, its n-factor is 1.
We know that, normality = molarity $\times $ n-factor
or, 0.3 = molarity $\times $ 1.
Therefore, the concentration is 0.3M.
Now we know that-
 $\begin{align}
  & molarity=\dfrac{no.of\text{ moles of solute}}{vol.of\text{ solvent(litre)}} \\
 & or,no.of\text{ moles = molarity}\times \text{volume(litres)} \\
\end{align}$

Here, 20cc of NaOH was used i.e 20mL and 1mL is one thousandth of 1litre. So, 20ml = 0.002L
Therefore, number of moles of NaOH = $0.3\times 0.02=6\times {{10}^{-3}}moles$
Now, as we can know that HCl is a monobasic acid and so is NaOH. Therefore, it is 1:1 reaction i.e. the number of moles of HCl will be equal to the number of moles of NaOH.
Therefore, the number of moles of NaOH = $6\times {{10}^{-3}}moles$.
Initially, we had 50ml of 0.5N solution of HCl. The n-factor of HCl is also 1. Therefore normality will be equal to molarity of HCl too.

 Let’s calculate the number of moles of HCl initially.
Number of moles of HCl initially = 0.5$\times $0.05 =0.025 moles.
Initially we had 0.025 moles of HCl and after reaction we were left with 0.006 moles.
Therefore, the number of moles of HCl reacted = 0.025 – 0.006 = 0.019 moles.
Now, for the given question, we can write the reaction as-
     \[Ca{{(OH)}_{2}}+2HCl\to CaC{{l}_{2}}+2{{H}_{2}}O\]

As we can see that the above reaction is a 2:1 reaction. Therefore, to find the number of moles of calcium hydroxide, we have to divide it by 2.

Therefore, the number of moles of calcium hydroxide = $\dfrac{0.019}{2}=9.5\times {{10}^{-3}}$.
Now, to find the mass of calcium hydroxide, we have to multiply the number of moles by molar mass.
Molar mass of calcium hydroxide = 74.

Therefore, the mass of calcium hydroxide = $9.5\times {{10}^{-3}}\times 74$= 0.703g.
We know that, \[Percentage\text{ purity=}\dfrac{Mass\text{ obtained}}{Mass\text{ of sample}}\times 100\]

Therefore, putting the mass of the pure sample of calcium hydroxide added and the mass of impure calcium hydroxide, we get-
     \[Percentage\text{ purity=}\dfrac{0.703}{50}\times 100=1.406%\]
Therefore, the purity of $Ca{{(OH)}_{2}}$ is 1.406%

Note: We know that we use back titration in volumetric analysis to find out the percentage purity of a sample as we did in the above question. We also use it to find out the excess of reagent added by titrating it with a suitable reagent. In an acid – base titration, if we have added excess base in the acid mixture, we can titrate the base with another acid of known strength to find out the excess base added.