A 50.0-mL volume of 0.15M HBr is titrated with 0.25M KOH. Calculate the pH after the addition of 18.00mL of KOH

Question

anonymous · Answer

meq of HBr=50*0.15=7.5
meq of KOH=0.25*18=4.5
meq of HBr left=3.
conc=3*10^-3 pH= 3-log 3=3-0.477=2.523

anonymous · Answer

thanks..but could you explain to me please?meq = Molarity at equilibrium?

anonymous · Answer

meq=milli equivalents. which in this case is just milli moles.

anonymous · Answer

oh I see...that is why you don't write m with capital letter?
but,the real answer is pH= 1.36

anonymous · Answer

Ah ok i got my mistake i took the concentration wrong :) 
moles=3*10^-3 volume=68.
so conc=3*10^-3/68. Now take -log of that.

anonymous · Answer

oh..thanks .I figured it out :D