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Chemistry 9 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

oh..thanks .I figured it out :D

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