I am an IB student. And I am trying to find the mass percentage of $\ce{NaHCO3}$ in antacid. I am dissolving the antacid in $\pu{0.1 M}$ $\ce{HCl}$ solution. So, first of all I conducted some preliminary experiments with pure $\ce{NaHCO3}$ powder to optimize the experiment, get a hang of it etc.

For this preliminary experiment, where I used $\pu{0.5 g}$ $\ce{NaHCO3}$, I calculated the amount of $\ce{HCl}$ solution required as being approx. $\pu{60 ml}$. Hence I added $\pu{100 ml}$ in the flask so that I have an excess of the acid as required.

Is adding $\pu{40 ml}$ as the supposed excess amount reasonable? Should it be more or less? I am confused because my teacher suggested that by even putting $\pu{100}$ ml of $\ce{HCl}$ in $\pu{0.5 g}$ of antacid tablets, I still don't have the acid in excess? But do I not? Doesn't it only take $\pu{60 ml}$ to neutralize $\pu{0.5 g}$ of $\ce{NaHCO3}$? So $\pu{100 ml}$ is definitely in excess, isn't it?


1 Answer 1


The volume you've established seems correct, as well as your assumption that extra $\pu{40 mL}$ of $\ce{HCl}$ would be a good excess:

$$\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}$$

$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{C(\ce{HCl})} = \frac{m(\ce{NaHCO3})}{M(\ce{NaHCO3}) \cdot C(\ce{HCl})} = \frac{\pu{0.5 g}}{\pu{84.0 g mol-1} \cdot \pu{0.1 mol L-1}} = \pu{59.5 mL}$$

I suspect your teacher either prepared a series of various antacids, or this one contains additional components, such as $\ce{CaCO3}$, $\ce{Mg(OH)2}$ of $\ce{Al(OH)3}$. Note that for the neutralization of the same mass of, say, aluminium(III) hydroxide, over than 3 times more $\ce{HCl}$ is required:

$$\ce{Al(OH)3 + 3 HCl -> AlCl3 + 3 H2O}$$

$$V(\ce{HCl}) = \frac{n(\ce{HCl})}{C(\ce{HCl})} = \frac{3m(\ce{Al(OH)3})}{M(\ce{Al(OH)3}) \cdot C(\ce{HCl})} = \frac{3 \cdot \pu{0.5 g}}{\pu{78.0 g mol-1} \cdot \pu{0.1 mol L-1}} = \pu{192 mL}$$

Either way, it looks like you understand how to properly find out the amount of acid for the neutralization. I think if you demonstrate your detailed attempt to the teacher, it should clarify the situation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.