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When two reactants partially react, and we are asked about concentration of one reactant, should we consider the volume of reactants, or if the products?

Here's an question to make it clear.

How much volume of $0.2$ M $\ce{H2SO4}$ to be added in $40$ mL of $0.1$ M NaOH so that resulting solution has concentration of 6/55 M NaOH?

The balanced reaction would be, $$ \ce{H2SO4 + 2 NaOH → Na2SO4 + 2 H2O} , $$ where all reactants and products are aqueous.

In brief, here's what sources say,

  1. Let $V$ ml of $\ce{H2SO4}$

  2. Considering millimoles, $0.2$ V millimoles of $\ce{H2SO4}$ react with 0.2 millimoles of NaOH.

  3. $0.2V - 0.2/ V + 40 = 6/55$

Solving, $V = 70$ mL.

However, they considered the volume of reactants, why not take into account volume of products?

Please clarify. Thanks.

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  • $\begingroup$ As it is an additional question about your prior query, it would be better to extend your original question, or provide reference to it, as I have just done. $\endgroup$
    – Poutnik
    Commented Aug 26 at 5:04
  • $\begingroup$ BTW, with volume, you'll need to consider the density of reactants. Mass makes it easy to calculate molarity, but will be slightly off 1 g/cm even at 0.2 M H2SO$, for example. $\endgroup$
    – DrMoishe Pippik
    Commented Aug 26 at 5:36
  • $\begingroup$ The sources used 0.2V - 2, not 0.2V - 0.2. // There is considered volume of solutions, not volume of reactants nor products. V for H2SO4 solution, 40 mL for NaOH solution, V + 40 mL for resulting solution. $\endgroup$
    – Poutnik
    Commented Aug 26 at 5:54
  • $\begingroup$ @DrMoishePippik While you are obviously right and total volume is not exactly 40 + V mL, the task assumes simplification ignoring density changes and is already above the head of the student. :-) // It is possible to find tabelated densities for H2SO4 and NaOH solutions, but not for mixed Na2SO4 + H2SO4 solutions. $\endgroup$
    – Poutnik
    Commented Aug 26 at 6:22
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    $\begingroup$ The original question wanted 6/55M excess H2SO4 not NaOH. $\endgroup$
    – jimchmst
    Commented Aug 27 at 4:42

2 Answers 2

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You have probably on your mind that we should not use as the final volume

$$V_\text{fin} = V_\ce{H2SO4(aq)} + V_\ce{NaOH(aq)}.$$

Exactly evaluated, you are right. As

$$V_\text{fin} = \frac{m_\text{fin}}{\rho_\text{fin}} = \frac{V_\ce{H2SO4(aq)}\cdot \rho_\ce{H2SO4(aq)} + V_\ce{NaOH(aq)}\cdot \rho_\ce{NaOH(aq)}}{\rho_\text{fin}}$$

The problem is that we can obtain tabulated density data for solution of $\ce{NaOH}$ and $\ce{H2SO4}$ but not of mixed solution of $\ce{H2SO4}$ and $\ce{Na2SO4}$. Aside of that, the task does not provide any density data.

It is obvious the task creator applies the simplification of volume additivity, that is not generally valid, just as an approximation. For such diluted solution, the error is small.

$$V_\text{fin} = V_\ce{H2SO4(aq)} + V_\ce{NaOH(aq)}$$

$$ \rho_\text{fin} = \frac{m_\ce{H2SO4(aq)} + m_\ce{NaOH(aq)}}{V_\ce{H2SO4(aq)} + V_\ce{NaOH(aq)}} $$

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Since it appears you asked almost the same question twice and got lots of math answers, I am going to try to answer more generally, to your conceptual question. Think in moles, not volume. You know your target concentration, 6/55 M, but you do not know your total final volume because the more H2SO4 you add, the more final volume you have. But you know how many moles of NaOH you start with, and you can write an expression for how many you must end with (concentration x volume). You must determine how much of the reactant to add, and as you add it you are both diluting the NaOH and reacting away the OH. So you again cannot simply determine how many moles of H2SO4 you must add, since you don't know how many moles of OH you need. You have to make an expression for that as well. In conclusion, the volume of reactant solution is what you must find, but you must include the volume of the final solution in your calculation.

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