Question

How many moles of sodium hydroxide would react with 1 Mole of sulphuric acid?

Answer 1

Answer:

Two moles.

Explanation:

Sulphuric (sulfuric) acid $\rm H_2SO_4$ is a diprotic acid. When one mole of $\rm H_2SO_4$ molecules dissolve in water, two moles of $\rm H^{+}$ ions would be produced.

$\rm H_2SO_4 \to 2\, H^{+} + {SO_4}^{2-}$.

On the other hand, sodium hydroxide $\rm NaOH$ is a monoprotic base. When one mole of $\rm NaOH$ formula units dissolve in water, only one mole of hydroxide ions $\rm OH^{-}$ would be produced.

$\rm NaOH \to Na^{+} + OH^{-}$.

Note that $\rm H^{+}$ and $\rm OH^{-}$ react at a one-to-one ratio:

$\rm H^{+} + OH^{-} \to H_2O$.

As a result, it would take $2\; \rm mol$ of $\rm OH^{-}$ to react with the $\rm 2\; mol$ of $\rm H^{+}$ that was released when $1\; \rm mol$ of $\rm H_2SO_4$ is dissolved in water. Since one mole of $\rm NaOH$ formula units could produce only one mole of $\rm OH^{-}$, it would take $\rm 2\; mol$ of $\rm NaOH$ formula units to produce that $2\; \rm mol$ of $\rm OH^{-}$ for reacting with $1\; \rm mol$ of $\rm H_2SO_4$.