Question: 2.5 g of the carbonate of metal was treated with 100ml of 1 N \[{H_2}S{O_4}\] . After the completion...

Question

2.5 g of the carbonate of metal was treated with 100ml of 1 N ${H_2}S{O_4}$ . After the completion of the reaction, the solution was boiled off to expel $C{O_2}$ and was then titrated against 1 N NaOH solution. The volume of alkali that would be consumed, if the equivalent weight of the metal is 20 is:
A. 50
B. 25
C. 75
D. 100

Answer 1

Solution

In titration the solution of a substance of unknown concentration is made to react with another solution whose concentration is known, so that the concentration of the substance can be found. Another important concept is of equivalent weight; it is given by weight of solute /number of gram equivalents.

Complete step by step answer:
in such cases if the normality of the substance whose concentration is known is N1 and its volume is V1 then its grams equivalents will be N1V1. According to the law of equivalents the gram equivalents of the substance with unknown concentration will be NV and if its volume is V its normality will be given by (N1*V1)/V. from the given data in the question we can conclude that
equivalent weight of $CO_3^{2 - } =$ atomic mass/valence factor ( where atomic mass of carbon is 12 and oxygen is 16.)
Equivalent weight of carbonate $CO_3^{2 - } =$ $\dfrac{{12 + (3 \times 16)}}{2} = \dfrac{{60}}{2} = 30$
Equivalent weight of metal carbonate = 20 + 30= 50.( since we are given with the equivalent weight of metal therefore it will be added to the weight of calculated carbonate. Further we are initially provided with 2.5 g of metal carbonate so the equivalent of the given 2.5g will be)
2.5 g of metal carbonate = $\dfrac{{2.5}}{{50}} = 0.05$ eq
On titration we find that
Number of equivalent of ${H_2}S{O_4}$ would have reacted = 0.05 eq
Equivalent of ${H_2}S{O_4}$ taken( NX V) = 1 N X 100 X ${10^{ - 3}}$ = 0.1 eq
Number of equivalent of ${H_2}S{O_4}$ remains unreacted = 0.1 - 0.05 = 0.05 eq
Number of equivalent of NaOH consumed = 0.05 eq
We know from law of equivalents that
Equivalent of NaOH = Eq. of H2SO4
Milli equivalents = N X V (in mL)
1 x V = 0.05 x 1000
V (in mL) = $\dfrac{{0.05 \times 1000}}{{1.0}} =$ 50 ml

So, the correct answer is “Option A”.

Note:
equivalent weight has to be carefully calculated, and the formulas to calculate equivalent weights. In the beginning we see valence factor for salts valence factor can be calculated by
n=number of moles of cation in salt*oxidation state of the cation.