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# $30ml$of $\dfrac{N}{{10}}HCl$ is required to neutralize $50ml$ of sodium carbonate solution. How many $ml$ of water must be added to $30ml$ of this solution. So that the solution obtained may have a concentration equal to $\dfrac{N}{{50}}?$

Last updated date: 13th Sep 2024
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Hint: $HCl$ acid reacts with sodium hydro carbonate to form sodium chloride and hydrogen carbonate. First find the normality of sodium carbonate solution. Then use that to find the volume of the new solution added to make the normality equal to $\dfrac{N}{{50}}$

Complete Step by step answer:It is given in the question that,
Volume of HCl, ${V_{HCl}} = 30ml$
Normality of HCl, ${N_{HCl}} = \dfrac{N}{{10}}$
$\Rightarrow {N_{HCl}} = 0.1N$
Volume of sodium carbonate, ${V_{N{a_2}C{O_3}}} = 50ml$
Given volume of HCl is neutralizing the given volume of sodium carbonate solution. Therefore, we can write,
${V_{N{a_2}C{O_3}}} \times {N_{N{a_2}C{O_3}}} = {V_{HCl}} \times {N_{HCl}}$
$\Rightarrow {(V \times N)_{N{a_2}C{O_3}}} = {(V \times N)_{HCl}}$
Now, by substituting the values given in the question, we can write,
${(50 \times N)_{N{a_2}C{O_3}}} = {(30 \times 0.1N)_{HCl}}$
$\Rightarrow 50 \times {N_{N{a_2}C{O_3}}} = 30 \times 0.1$
Rearranging it we can write
${N_{N{a_2}C{O_3}}} = \dfrac{{30 \times 0.1}}{{50}}$
On simplifying it, we get
$\Rightarrow {N_{N{a_2}C{O_3}}} = \dfrac{3}{{50}}$
$\Rightarrow {N_{N{a_2}C{O_3}}} = 0.06N$
Now, we have to add some solution to $30ml$ of $0.06N$, $N{a_2}C{O_3}$ such that its normality is $\dfrac{N}{{50}}$
$\dfrac{N}{{50}} = 0.02N$
We will again use the formula,
${N_1}{V_1} = {N_2}{V_2}$
Where,
${N_1}$ is the previous normality of the sodium carbonate solution
${V_1}$ is the given volume of the sodium carbonate solution
${N_2}$ is the new expected normality of sodium carbonate solution
${V_2}$ is the volume that we need to add in the solution
By rearranging the above formula, we can write
${V_2} = \dfrac{{{N_1}{V_1}}}{{{N_2}}}$
Putting the values we get,
$\Rightarrow {V_2} = 90ml$

Therefore, $90ml$ of water should be $X$ added to get $0.02N$ solution.

Note: Neutralize hydrochloric acid with an alkali.Such as sodium bicarbonate. Wearing your protective garments and working in a ventilated area well away from children, pills, heat and metals prepare a base min, min $1/b$of baking soda with plenty of water, slowly add the hydrochloric acid.