How much additional hydrochloric acid should be added to arrive at ‘excess acid’?

Question

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?

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.