Stoichiometry of Neutralisation Reaction

Question

For the following question, though the answers are available on web, none of them are providing exact reason behind WHY the specific formula was used.

Question -

What Volume (in $\pu{mL}$) of $\pu{0.2 M}\ \ce{H2SO4}$ solution should be mixed with the $\pu{40 mL}$ of $\pu{0.1 M}\ \ce{NaOH}$ solution such that the resulting solution has the concentration of $\ce{H2SO4}$ as $\pu{\frac{6}{55} M}$.

Approach and Confusions:

1. This is a neutralisation neaction. So I wrote balanced equation as,

   $$\ce{2 NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + 2 H2O(l)}$$

2. As per the question, formed solution contains $\pu{\frac{6}{55} M}\ \ce{H2SO4}$. Hence, I considered that originally, amount in moles of $\ce{H2SO4}$ must be greater. I wrote the equation as,

   $$\ce{2x NaOH(aq) + ($x+y$) H2SO4(aq) -> y H2SO4(aq) + x Na2SO4(aq) + 2x H₂O(l)}$$

3. From the question, $\dfrac{y\ \ce{H2SO4} }{ V [\text{L}]} = \pu{\dfrac{6}{55} M}$

I considered volume of $\ce{Na2SO4}$ to be negligible as it's solid, and without this assumption I was unable to proceed further. [editor note: will be dissolved]

4. Again from the question, $\pu{40 mL}\ \pu{0.1 M}\ \ce{NaOH} = \pu{0.004 mol} = 2x$.

After this I wasn't able to calculate the exact value of volume $\pu{0.2 M}\ \ce{H2SO4}$ which was $\pu{70 mL}$.

Please assist about the correct answer. It would be helpful if you analysed my approach and told where or why it's not working.

Answer 1

I choose another approach, to get the same final solution : $70$ mL of $0.2$ M $\ce{H2SO4}$.

I divide the problem in two parts : First neutralize precisely the amount of $\ce{NaOH}$ and then add enough $\ce{H2SO4}$ to the neutral solution to get a final $\ce{H2SO4}$ concentration equal to $6/55$ M.

First part : Neutralization. The initial solution is $40$ mL $0.1$ M $\ce{NaOH}$. It contains then $4.0$ millimol $\ce{NaOH}$. This needs $2.0$ millimol $\ce{H2SO4}$ to be neutralized. This required amount is included in a volume $\pu{V_o}$ of $0.2$ M $\ce{H2SO4}$. $$\ce{V_o = \frac{n}{c} = \frac{2 mmol}{0.2mmol/L}= 10 mL}$$ After this operation the volume of the solution is $40 + 10 = 50$ mL, and it is a solution of $\ce{Na2SO4}$ .

Second part : Dilution of acid by adding a volume $\ce{V_1}$ of $0.2$ M $\ce{H2SO4}$ (containing $\ce{n_1}$ moles $\ce{H2SO4}$) to $50$ mL of the then neutral solution of $\ce{Na2SO4}$. The final concentration $\ce{c_f}$ = $\ce{\frac{6}{55}}$ is given by : $$\ce{c_f = \frac{n_1}{V_1 + 0.05} = \frac{V_1* 0.2}{V_1 + 0.05} = \frac{6}{55}}$$ $$\ce{V_1* 0.2 = \frac{6}{55}(V_1 + 0.05)}$$ After multiplication on both sides by $11$ : $$\ce{2.2*V_1 = \frac{6}{5} (V_1 + 0.05) = 1.2 V_1 + 0.06}$$ This corresponds to an addition of $\ce{V_1 = 0.06 L = 60 mL}$ of $0.2$ M $\ce{H2SO4}$ .

The total volume of used $0.2$ M $\ce{H2SO4}$ is :

$$\ce{V_o + V_1 = 10 mL + 60 mL = 70 mL}$$

Answer 2

Using $\pu{x mL}$ of $\pu{0.2 M}\ \ce{H2SO4}$ brings $0.2x\ \pu{mmol}$ of $\ce{H2SO4}$.

There is spend $\frac{\pu{40 mL} \times \pu{0.1 mmol mL-1}}{2}\ = \pu{2 mmol}$ of $\ce{H2SO4}$.

There is remaining $0.2x\ \pu{mmol} -\pu{ 2 mmol}$ of $\ce{H2SO4}$,

of molar concentration $$\dfrac{0.2x\ \pu{mmol} -\pu{ 2 mmol}}{40\ \pu{mL} + x\ \pu{mL}}=\dfrac{0.2x - 2 }{40 + x} \ \pu{mol L-1} = \pu{\frac{6}{55} mol L-1}$$

this leads to a simple linear algebraic equation:

$$0.2x - 2 = (40 + x)\frac{6}{55}$$

$$x(0.2 - \frac{6}{55}) = \frac{240}{55} + 2$$

$$x(\frac{11-6}{55}) = \frac{240+110}{55}$$

$$5x = 350$$

$$x = 70$$

The required volume of $\pu{0.2 M}\ \ce{H2SO4}$ is then $\ce{70 mL}$.

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Another approach: there is spent $\ce{10 mL}$ of $\ce{H2SO4}$ solution on $\pu{40 mL}$ of $\ce{NaOH}$ solution, it makes 50 mL of liquid without any $\ce{H2SO4}$.

Then by the rule the concetration is reciprocal to dilution. It can be seen also as the equivalent task - what volume of $\pu{0.2 M}\ \ce{H2SO4}$ should be taken to get concentration $\pu{\frac{6}{55} M}$ by diluting it by $\ce{50 mL}$ of water:

$$\frac{x-10}{x+40} = \dfrac{(\frac{6}{55})}{(\frac{11}{55})} = 6/11$$

$$(x-10)11 = 6(x+40)$$

$$5x = 240+110=350$$

$$x = 70$$

The required volume of $\pu{0.2 M}\ \ce{H2SO4}$ is then $\pu{70 mL}$.