Question

You use 25mL of 6.0M NaOH to neutralize 100mL of an unknown acid in a titration experiment. What is the molarity of the acid? Show work

Answer 1

Problem: You use 25mL of 6.0M NaOH to neutralize 100mL of an unknown acid in a titration experiment. What is the molarity of the acid? Show work.

Solution: The shortened ionic chemical equation of neutralization: $\mathrm{H}^{+} + \mathrm{OH}^{-} = \mathrm{H}_{2}\mathrm{O}$;

The quantities of moles of $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ are equal: $\nu (\mathrm{H}^{+}) = \nu (\mathrm{OH}^{-})$; quantity can be obtained from the following expression for concentration $c = \nu / V$: $\nu = c^{*}V$ (where $c$ - concentration; $V$ - volume).

So, $c(\mathrm{H}^{+} ) * V(\mathrm{H}^{+}) = c(\mathrm{OH}^{-}) * V(\mathrm{OH}^{-})$; $c(\mathrm{H}^{+}) = c(\mathrm{OH}^{-}) * V(\mathrm{OH}^{-}) / V(\mathrm{H}^{+}) = 6.0 * 25 / 100 = 1.5$ (M).

Answer: $c(\mathrm{H}^{+} ) = 1.5\mathrm{M}$