# Unit analysis for neutralization reaction

What volume of $$\pu{0.500 mol L-1}$$ $$\ce{HCl}$$ (in $$\pu{mL}$$) is required to neutralize $$\pu{2.02 g}$$ of solid $$\ce{NaOH}$$ pellets?

So far I have tried regular unit analysis:

2.02 g NaOH * 1mol NaOH/40g NaOH * 1mol HCl/1mol NaOH * 500g HCl/1mol HCl = 25.25 g HCl --> 25.5 mL HCl

This answer doesn't make sense in the context of this neutralization reaction. Is there a type of unit analysis that is used specifically for determining the volume required for a reaction?

• Could you explain your calculations? – William R. Ebenezer Apr 5 '19 at 12:43

There is no special unit analysis. It's a recommended practice to solve the problem algebraically first using proper notations for physical quantities, and plug the numeral values at the end minding the units — this way you reduce the chance of making the erroneous calculations and keep track of all units; as a bonus, you simplify handling significant figures.

The majority of problem solving in chemistry begins with writing down the balanced chemical reaction:

$$\ce{HCl + NaOH → NaCl + H2O}$$

So, to find the volume of hydrochloric acid $$V(\ce{HCl})$$, you may use its molar concentration $$c(\ce{HCl})$$ and the amount $$n(\ce{HCl})$$:

$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{c(\ce{HCl})}$$

Unknown amount $$n(\ce{HCl})$$ can be found from the reaction's stoichiometry:

$$n(\ce{HCl}) = n(\ce{NaOH}) = \frac{m(\ce{NaOH})}{M(\ce{NaOH})}$$

where $$m(\ce{NaOH})$$ and $$M(\ce{NaOH})$$ are mass and molecular mass of sodium hydroxide, respectively. Putting everything together and plugging in the appropriate numerical values, you arrive at the volume of approx. $$\pu{100 mL}$$:

\begin{align} V(\ce{HCl}) &= \frac{m(\ce{NaOH})}{M(\ce{NaOH})\cdot c(\ce{HCl})} \\ &= \frac{\pu{2.02 g}}{\pu{40.00 g mol-1}\cdot \pu{0.500 mol L-1}} \\ &= \pu{0.101 L} \end{align}

The Unit/dimension analysis is in this case rather "using a cannon against sparrows".

The first spotted error is $$\pu{500 g}$$ of $$\ce{HCl}$$. Where did it come from ?

Note that the molar mass of $$\ce{HCl}$$ is about $$\pu{36.5 g / mol}$$.

1. Get the molar mass of $$\ce{NaOH}$$.
2. Calculate the amount of mols of $$\ce{NaOH}$$.
3. Calculate the equivalent amount of mols of $$\ce{HCl}$$.
4. Calculate the equivalent volume of $$\pu{0.500 mol / l}$$ $$\ce{HCl}$$.
• For a beginning chemistry student using dimensional analysis is a good thing. – MaxW Apr 5 '19 at 15:43
• I agree. If he did not introduce those 500 g, he could land at the correct result with the analysis as an added value.. – Poutnik Apr 5 '19 at 15:54