# Neutralization of alkali solution by a mix of strong acids

$\pu{0.1 M}$ $\ce{HCl}$ and $\pu{0.2 M}$ $\ce{H2SO4}$ solutions are mixed in equal volume. This solution is diluted to double the volume. Find the volume of $\pu{0.1 M}$ $\ce{NaOH}$ which can be neutralized with $\pu{20 mL}$ of the former solution.

The answer is $\pu{50 mL}$.

I have tried a bit, but the answer I get is wrong. Let volume be $V$ at the beginning, therefore, no. of moles of $\ce{HCl}$ is $0.1V~\pu{mol}$, no. of moles of $\ce{H2SO4}$ is $0.2V~\pu{mol}$.

Total volume of solution after dilution is $4V$, therefore, no. of moles of $\ce{H+}$ in the solution is $(0.1V + 0.4V)/4V$.

No. of moles of $\ce{Cl-}$ in the solution is $0.1V/4V = 1/40$.

No. of moles of $\ce{SO4^2-}$ in the solution is $1/20$.

Since $\ce{NaOH}$ reacts only with $\ce{Cl-}$ and $\ce{SO4^2-}$, therefore total moles of $\ce{NaOH}$ needed is $(1/20 + 1/40)(20/1000) = 1/100~\pu{mol}$

Therefore, $\text{volume needed} = \text{molarity/no. of moles} = 0.1/0.01 = \pu{10 L} = \pu{10000 mL}$.

I Don't Know what I am doing wrong.

I tried again using some of the suggestions given: for the first step of mixing H2SO4 and HCl, M1V+M2V=M3(2V), which comes as M3=0.5/2. Then for dilution, (0.5/2)*2V=M4*4V thus M4=0.5/4. Taking 20ml, (0.5/4)*20=0.1*x, [0.1M of NaOH] thus x=25ml. Why doesn't this work??

• Sodium hydroxide does not react with either chloride or sulphate, only with hydronium ions/protons.
– Jan
Dec 13 '17 at 1:20

I think you over-complicated things a bit by focusing on the formation of sodium salts, when in reality all you need to take into account is the net neutralization reaction:

$$\ce{H3O+ + OH- -> 2 H2O}$$

From the equivalence point condition of any neutralization reaction one can find volume of the base $V_\mathrm{b}$ expressed via the volume of acid(s) $V_\mathrm{a}$:

$$V_\mathrm{b} = \frac{C(\ce{H3O+})}{C(\ce{OH-})} \cdot V_\mathrm{a} \tag{1}$$

Since there are two moderately diluted strong acids and a base, one can conclude that the dissociation of each acid and a base is independent and complete:

\begin{align} \ce{HCl + H2O &<=>> H3O+ + Cl-} \\ \ce{H2SO4 + 2H2O &<=>> 2H3O+ + SO4^2-} \\ & \\ \ce{NaOH &<=>> Na+ + OH-} \end{align}

Note that concentration of protons is doubled in case of sulfuric acid. Total concentration of protons in the solution is

$$C(\ce{H3O+}) = \frac{n(\ce{H3O+})}{V(\ce{H3O+})} = \frac{n(\ce{HCl}) + 2 \cdot n(\ce{H2SO4})}{V(\ce{HCl}) + V(\ce{H2SO4})} \tag{2}$$

Taking into account that $n = CV$ and that two identical volumes of acid were mixed ($V(\ce{HCl}) = V(\ce{H2SO4}) = V(\text{acid})$) (2) can be transferred into:

\require{cancel} \begin{align} C(\ce{H3O+}) &= \frac{C(\ce{HCl}) \cdot V(\ce{HCl}) + 2 \cdot C(\ce{H2SO4}) \cdot V(\ce{H2SO4})}{V(\ce{HCl}) + V(\ce{H2SO4})} \\ &= \frac{\cancel{V(\text{acid})} \cdot (C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4}))}{2 \cdot \cancel{V(\text{acid})}} \\ &= \frac{C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4})}{2} \tag{3} \end{align}

Now let's plug (3) in (1) and finally find the volume of alkali:

$$V_\mathrm{b} = \frac{C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4})}{2 \cdot C(\ce{OH-})} \cdot V_\mathrm{a} = \frac{\pu{0.1 M} + 2 \cdot \pu{0.2 M}}{2 \cdot \pu{0.1 M}} \cdot \pu{20 mL} = \pu{50 mL}$$

– Sri
Dec 13 '17 at 10:21
• @Sri To find the molar concentration of any compound ($\ce{H3O+}$ in this case), we need to divide amount by volume. Since we mix two acids, the resulting volume is the sum of their volumes. Dec 13 '17 at 10:29
• And where have we used the fact that the volume has been doubled??
– Sri
Dec 14 '17 at 10:55
• @Sri In divisor in eq. (2) we add two equivalent volumes of both acids, isn't that it? Dec 14 '17 at 11:12
• @Sri I expanded the expression for the concentration of protons so that now it should be completely transparent what is what. Dec 14 '17 at 11:32

First, thank you for showing work for a homework problem. I think it is instructive in this case, not only to provide the correct solution, but also to point out where your problem solving went wrong. The exact solution to this problem may not be extensible to other problems, but the general guidelines for solving problems will, so if there's a take-home message, those are it.

1. You're talking about dilution. Why does that matter here? The crux to any problem is to figure out what exactly are the key features you need to extract. This problem requires understanding stoichiometry because you need to match amounts of a substance according to a reaction and concentration because you need to understanding the relationship between amount of substance, volume, and concentration.
2. Units. You have units in some places but not others. Units should be carried through the entire calculation. What is 100? It's a number. It's not a measurement or even a quantity because it has no units. Having units is a check on performing the correct calculations because units will cancel when dividing/multiplying and only two values with the same units can be added.
3. Sanity checking results. 10 mL is roughly a small mouthful of water. 10000 mL is the volume of a large pot. For two solutions of roughly the same concentration, there should not be a 1000x different in the amounts.
4. Make sure you have what you need before you start solving. You correctly tagged this as a stoichiometry problem. The correct stoichiometric ratios are $\ce{HCl}:\ce{NaOH} = 1 : 1$ and $\ce{H2SO4}:\ce{NaOH} = 1 : 2$, but even if you had said that, you don't have the means to justify it. Why? Because stoichiometry starts with balanced equations which you don't have. Once you determine the type of problem, make sure you have the appropriate ground work for solving problems of that type.

OK. So onto solving your problem. For point 4, you should write out balanced equations:

$$\ce{HCl + NaOH -> H2O + NaCl}$$ $$\ce{H2SO4 + 2NaOH -> 2H2O + Na2SO4}$$

At this point, you immediately know that you can decompose the problem into two separate problems. Why? Because hydrochloric acid and sulfuric acid don't interact with each other according to the equations written.

At this point, we should compute the total moles of sodium hydroxide required since this is the whole point of stoichiometry.

Note that since the initial solutions are mixed in equal volume, the concentration of each is half what is listed. This is definitely a dilution problem, but one that is simple enough not to require any kind of dilution math or ratios aside from dividing by 2.

$$0.02\ \mathrm{L}\cdot\frac{0.05\ \mathrm{mol\ \ce{HCl}}}{L}\cdot\frac{\mathrm{mol\ \ce{NaOH}}}{\mathrm{mol\ \ce{HCl}}}= 0.001\ \mathrm{mol\ \ce{NaOH}}$$

$$0.02\ \mathrm{L}\cdot\frac{0.1\ \mathrm{mol\ \ce{HCl}}}{L}\cdot\frac{2\ \mathrm{mol\ \ce{NaOH}}}{\mathrm{mol\ \ce{HCl}}}= 0.004\ \mathrm{mol\ \ce{NaOH}}$$

The total amount of sodium hydroxide is:

$$0.001\ \mathrm{mol\ \ce{NaOH}} + 0.004\ \mathrm{mol\ \ce{NaOH}} = 0.005\ \mathrm{mol\ \ce{NaOH}}$$

Now, concentration is the per unit volume amount of substance:

$$c = \frac{n}{V}$$

or

$$V = \frac{n}{c}$$

Plugging in:

$$V = 0.005\ \mathrm{mol\ \ce{NaOH}}\cdot\frac{L}{0.1\ \mathrm{mol\ \ce{NaOH}}} = 0.05\ \mathrm{L}$$

Now. Sanity check: 50 mL is much closer to 20 mL and makes more sense.

• Thanks, I don't quite get the first equation can you please explain it. It would be nice
– Sri
Dec 17 '17 at 13:27
• @Sri If you don't understand the first chemical equation, that's why you have problems. Please review neutralization. If you don't understand the first mathematical expression, that also would indicate why you have a problem. Please review the concept of concentration.
– Zhe
Dec 20 '17 at 19:37