Question

If 25.0 mL of NaOH was neutralised by 40.0 mL of 0.05 mol L-1 HNO3, the molarity of the NaOH is?

How do I work this out? I have seen some examples but it doesn't make any sense to me.

Answer 1

We address the stoichiometric equation:

`HNO_3(aq) + NaOH(aq) rarr NaNO_3(aq) + H_2O(l)`

Explanation:

And this establishes the 1:1 equivalence of nitric acid and sodium hydroxide....

Now `"molarity"="moles of solute"/"volume of solution"`

And thus `[NaOH]-="moles of nitric acid"/"initial volume of sodium hydroxide"`

`=(40.0*mLxx10^-3*L*mL^-1xx0.05*mol*L^-1)/(25.0*mLxx10^-3*L*mL^-1)`

`=0.080*mol*L^-1`..reasonably the sodium solution is MORE concentrated than the nitric acid solution given that a larger volume of acid was required for equivalence.