6
$\begingroup$

$55.0~\mathrm{mL}$ of $0.250~\mathrm{M}~\ce{NaOH}$ is used to titrate $35.0~\mathrm{ml}$ of $\ce{H2SO4}$. What is the molarity of $\ce{H2SO4}$?

I know that the equation for this reaction is: $$\ce{H2SO4 + 2NaOH <=> Na2SO4 + 2H2O}$$

I know that I have to use $$i_\mathrm a\cdot M_\mathrm a\cdot V_\mathrm a = i_\mathrm b\cdot M_\mathrm b\cdot V_\mathrm b$$ formula where $i$ is the number of ions, $M_\mathrm a$ and $M_\mathrm b$ are the molarities (of the acid and base, respectively), and $V_\mathrm a$ and $V_\mathrm b$ are the volumes of the acid and base (in milliliters).

I starting doing this problem and got up to this point:

$$i\cdot M_\mathrm a\cdot 35~\mathrm{ml} = 2\cdot 0.250~\mathrm{M}\cdot 55~\mathrm{ml}$$

So my question is: What should I put for the number of ions on the left hand side? Is it three or two?

$\endgroup$
  • $\begingroup$ You're almost right. $i_a$ is the number of $H^+$ ions and $i_b$ is the number of $OH^-$ ions. $\endgroup$ – LDC3 May 4 '14 at 21:17
6
$\begingroup$

That is one way of doing it.

In reality, this is a stoichiometric question where you are trying to find the molarity of an unknown polyprotic acid by titrating it with a strong base (woo-hoo!) of known molarity. I prefer using this method as it is easier to see the relationship between the units as well as use chain-ink conversion which most people are familiar with.

We know that

$$ \text{M} = \frac{\text{mol}}{\text{L}} $$

Therefore,

$$ 55 \ \text{mL} \ \ce{ NaOH} * \frac{250 \ \text{mmol} \ \ce{NaOH}}{1000 \ \text{mL} \ \ce{NaOH}} * \frac{1 \ \text{mmol} \ \ce{H2SO4}}{2 \ \text{mmol} \ \ce{NaOH}} * \frac{1}{ 35 \ \text{mL} \ \ce{H2SO4}} = $$

Which equals

$$ 0.19643 \ \frac{\text{mol}}{\text{L}} $$

$\endgroup$
  • 3
    $\begingroup$ In reality you would do this experiment probably with conductometry or for sulphate it is even better to do it via gravimetry via $$\ce{BaCl2 + H2SO4 -> BaSO4 v + 2 'HCl'}.$$ It should also be noted, that the second protolysis of suphuric acid does not behave like a strong acid. The equilibrium $$\ce{HSO4- (aq) + {}^{-}OH <=> H2O + SO4^{2-} (aq)}$$ is not fully on the right side. There is a significant amount of undissociated hydrogensulphate ions left (at pH=7). $\endgroup$ – Martin - マーチン May 5 '14 at 4:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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