How to determine the concentration of a base through indirect titration?

Question

$5.267~\mathrm{g}$ of $\ce{Na2CO3}$ was dissolved in $250.00\ \mathrm{mL}$ of water. $10.00\ \mathrm{mL}$ of this solution was titrated with $\ce{HCl}$. The end-point occurred at $21.30\ \mathrm{mL}$. This acid was titrated with $25.00\ \mathrm{mL}$ of $\ce{Ba(OH)2}$. The end-point occurred at $27.10\ \mathrm{mL}$. Calculate the concentration of the $\ce{Ba(OH)2}$. [Answer: $0.1013~\mathrm{M}$]

1. Using the mass they gave me ($5.267~\mathrm{g}$) I worked out the moles of $\ce{Na2CO3}$ which was $0.0497\ \mathrm{mol}$.

2. I then worked out the concentration of $\ce{Na2CO3}$ using the moles I just worked out and the volume they gave me ($0.25~\mathrm{L}$) and I got $0.1988~\mathrm{M}$.

3. The question says that only $0.01~\mathrm{L}$ was used, so I worked out the number of moles this would be which was $0.001988\ \mathrm{mol}$.

4. Using stoichiometry, I multiplied this value by two to get the moles of $\ce{HCl}$ that would have reacted with this, which was $0.003976\ \mathrm{mol}$.

5. It says that the end-point occurred at $21.30~\mathrm{mL}$, which if you subtract the volume of $\ce{HCl}$ which was already there, I worked out that $11.3~\mathrm{mL}$ or $0.0113~\mathrm{L}$ of $\ce{HCl}$ must have been used.

6. From here, I worked out the concentration of $\ce{HCl}$ (using $0.0113~\mathrm{L}$ and $0.003976\ \mathrm{mol}$) and got $0.35186~\mathrm{M}$.

7. It says that for the next reaction, $25~\mathrm{mL}$ of $\ce{Ba(OH)2}$ was used and the end-point occurred at $27.10~\mathrm{mL}$. From this information, I worked out that $0.0021~\mathrm{L}$ of $\ce{HCl}$ must have been used in this reaction.

8. By knowing the volume of $\ce{HCl}$ used, I worked out the amount (moles) of $\ce{HCl}$ used in the reaction, using the previously found concentration. I got $0.0007389\ \mathrm{mol}$.

9. Using stoichiometry, I worked out the amount of $\ce{Ba(OH)2}$ which must have reacted with this many moles of $\ce{HCl}$. I divided the $\ce{HCl}$ moles by $2$ and got $0.00036945\ \mathrm{mol}$.

10. By using the volume of $\ce{Ba(OH)2}$ used which they give me ($0.025~\mathrm{L}$) and the moles I just found ($0.00036945\ \mathrm{mol}$) I finally worked out the concentration of $\ce{Ba(OH)2}$, which was $0.0148~\mathrm{M}$.

My answer, $0.0148~\mathrm{M}$ is different to the answer the question gives ($0.1013~\mathrm{M}$), and I'm not sure why.

Answer 1

5. It says that the end-point occurred at $21.30\ \mathrm{mL}$, which if you subtract the volume of $\ce{HCl}$ which was already there, I worked out that $11.3\ \mathrm{mL}$ or $0.0113\ \mathrm{L}$ of $\ce{HCl}$ must have been used.

I think, this is where you went wrong. If a question says something along the lines of

The end-point occurs at $\dots~\mathrm{ml}$

then you should generally assume that that volume of whatever you are titrating with was used. In this case, that would mean a full $21.3~\mathrm{ml}\ \ce{HCl}$. The same thing of course happened at your point 7, where you write:

7. It says that for the next reaction, $25\ \mathrm{mL}$ of $\ce{Ba(OH)2}$ was used and the end-point occurred at $27.10\ \mathrm{mL}$. From this information, I worked out that $0.0021\ \mathrm{L}$ of $\ce{HCl}$ must have been used in this reaction.

Working through this bit by bit, and starting at your point 5:

$$c(\ce{HCl}) = \frac{n(\ce{HCl})}{V(\ce{HCl})} = \frac{0.003976~\mathrm{mol}}{0.0213~\mathrm{l}} = 0.1867~\mathrm{M}$$

For the titration of $\ce{Ba(OH)2}$:

$$n(\ce{HCl}) = c(\ce{HCl}) \times V(\ce{HCl}) = 0.1867~\mathrm{M} \times 27.10~\mathrm{ml} = 5.059~\mathrm{mmol} = 0.005059~\mathrm{mol}\\ n(\ce{Ba(OH)2}) = \frac{n(\ce{HCl})}{2} = \frac{5.059~\mathrm{mmol}}{2} = 2.529~\mathrm{mmol} = 0.002529~\mathrm{mol}\\ c(\ce{Ba(OH)2}) = \frac{n(\ce{Ba(OH)2})}{V(\ce{Ba(OH)2})} = \frac{2.529~\mathrm{mmol}}{25~\mathrm{ml}} = \frac{0.002529~\mathrm{mol}}{0.025~\mathrm{l}} = 0.1012~\mathrm{M}$$

The difference still there can be attributed to a rounding error somewhere along the way.

Answer 2

This is how I work out the answer which is 0.1012 M.