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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?

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  • $\begingroup$ Could you explain your calculations? $\endgroup$ – William R. Ebenezer Apr 5 at 12:43
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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. $\pu{100 mL}$:

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

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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 $\pu{0.500 mol / l}$ $\ce{HCl}$.
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  • 1
    $\begingroup$ For a beginning chemistry student using dimensional analysis is a good thing. $\endgroup$ – MaxW Apr 5 at 15:43
  • $\begingroup$ I agree. If he did not introduce those 500 g, he could land at the correct result with the analysis as an added value.. $\endgroup$ – Poutnik Apr 5 at 15:54

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