Question

If `25*cm^3` of `1.0*mol*L^-1` nitric acid are reacted with. `50*cm^3` `0.5*mol*L^-1` `NaOH` what is the `pH`?

Answer 1

`pH=7`

Explanation:

We interrogate the following acid-base reaction:

`HNO_3(aq) + KOH(aq) rarr KNO_3(aq) + H_2O`.

We use the old faithful formula:

`"Concentration"` `-=` `"Moles of solute"/"Volume of solution"`,

So with respect to nitric acid we have,

`25*cm^3xx10^-3*L*cm^-3xx1.0*mol*L^-1=0.025*mol`.

And with respect to potassium hydroxide we have,

`50*cm^3xx10^-3*L*cm^-3xx0.5*mol*L^-1=0.025*mol`

And thus there are equimolar quantities of nitric acid and sodium hydroxide, and at the end of addition we have a `75*cm^3` volume that is stoichiometric in potassium nitrate. And since potassium nitrate does not affect solution `pH`, we have a NEUTRAL solution, `pH=7`.