Titration of sulfuric acid with sodium hydroxide $55.0~\mathrm{mL}$ 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:
$$\ce{H2SO4 + 2NaOH &lt;=&gt; Na2SO4 + 2H2O}$$

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?
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 

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