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#?

1 Answer
Mar 29, 2017

Answer:

#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#.