Question

(a) 2 g of a metal carbonate were dissolved in 50 ml of N HCl. 100 ml of 0.1 N NaOH were required to neutralise the resultant solution. Calculate the equivalent mass of the metal carbonate.
(b) How much water should be added to 75 ml of 3 N HCl to make it a normal solution?

Answer 1

For (a): First calculate the volume of HCl to be neutralised by NaOH. Then from the remaining volume of HCl calculate the number of moles of HCl that neutralise the carbonate. Using these moles of HCl find the moles of carbonate involved, from which we can find out the molecular weight of the metal carbonate and thus its equivalent weight also.
For (b): Find the final volume of the solution and then subtract the initial volume to find the amount of water to be added.

Formula used:
-We will be using the dilution formula of normality:
${N_1}{V_1} = {N_2}{V_2}$ (1)
Where, N = normality and V is volume
-Molarity = Number of moles of solute / Volume of solution (in L) (2)
-Number of moles = given weight / molecular weight (3)
-Equivalent weight = Molecular weight / n-factor (4)

Complete step by step solution:
(a)- According to the question the solution formed by this mixing was neutralised by 0.1 N, 100 ml NaOH.
First we will find out the volume of 1 N HCl that is required for 0.1 N and 100 ml to neutralise using equation (1): ${N_1}{V_1} = {N_2}{V_2}$
Here, ${N_1}$ = Normality of NaOH solution = 0.1 N
${V_1}$ = Volume of NaOH solution = 100 ml
${N_2}$ = Normality of HCl solution = 1 N
${V_2}$ = volume of HCl solution
0.1×100 = 1× ${V_2}$
So, ${V_2}$ = 10 ml
So, the 10 ml of HCl will be required.
-Now let us see after 10 ml of HCl being neutralised, the amount of HCl left is = 50 – 10 = 40 ml or 0.04 L
-Concentration of HCl is given in the question to be 1 N. For HCl the n-factor is 1, so its equivalent weight is equal to its molecular weight and thus for HCl: 1 M = 1 N
From equation (2): M = moles of solute / Vol. of solution
1 M = moles / 0.04
Moles = 1×0.04 = 0.04 moles of HCl
This means that 0.04 moles of HCl will be used in the neutralisation of the carbonate.
-This neutralisation reaction of carbonate can be written as:
${M_2}C{O_3} + 2HCl \to 2MCl + {H_2}C{O_3}$
From this reaction we can see that 1 mole of ${M_2}C{O_3}$ is neutralised by 2 moles of HCl.
So, if 0.04 moles of HCl are used then the amount of ${M_2}C{O_3}$ neutralised will be:
Moles of ${M_2}C{O_3}$ = 0.04 /2 = 0.02 moles of ${M_2}C{O_3}$
-Given weight of ${M_2}C{O_3}$ is = 2 g and the moles we have calculated to be 0.02 moles.
So, we can calculate the molecular mass of ${M_2}C{O_3}$ from equation (3):
No. of moles = Given weight / Molecular weight
0.02 = 2 / Mol. Wt.
Mol. Wt. = 2 / 0.02 = 100 g
So, the molecular weight of ${M_2}C{O_3}$ is 100 g.
Since for all carbonates the n-factor is = 2 and so we calculate the equivalent weight of ${M_2}C{O_3}$ from equation (4): Equivalent weight = Molecular weight / n-factor
Eq. wt. = 100 / 2 = 50
So, the equivalent weight of the metal carbonate is 50.

(b)-The question gives us the initial volume and the initial concentration of HCl and we need to find out how much water has to be added in it to make a normal HCl solution.
We use equation (1) to calculate the final volume:
${N_1}{V_1} = {N_2}{V_2}$
Where, ${N_1}$ = normality of initial solution = 3 N
${V_1}$ = volume of initial solution = 75 ml
${N_2}$= normality of final solution after dilution = 1 N
${V_2}$= volume of final solution after dilution
3×75 = 1×${V_2}$
${V_2}$= 225 ml
-Now we have the final volume of the solution and to calculate the amount of water to be added we need to subtract the initial volume from the final volume.
Amount of water to be added = ${V_2} - {V_1}$
= 225 – 75
= 150 ml
So, the amount of water to be added to make the solution a normal solution is 150 ml.

Note:
For (a): In this question most of us get stuck at the concentration of HCl part. So, always remember that since HCl has an n-factor of 1, it’s molecular and equivalent weights are equal and so its molarity is equal to its normality.
For (b): Here we make a common mistake of taking ${V_2}$ to be the volume of water to be added. So, remember to subtract ${V_1}$ from ${V_2}$ to calculate the amount of water to be added.