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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
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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
thanks..but could you explain to me please?meq = Molarity at equilibrium?
meq=milli equivalents. which in this case is just milli moles.
oh I see...that is why you don't write m with capital letter? but,the real answer is pH= 1.36
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.
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oh..thanks .I figured it out :D
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