# Chemistry

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A 30.00mL sample of 0.150M KOH is titrated with 0.125M HClO4 solution. Calculate the pH after the following volumes of acid have been added: 30.0 mL, 35.0 mL, 36.0 mL, 37.0 mL, 40.0 mL

• Chemistry - ,

This is a strong acid/strong base titration so the pH at the equivalence point will be 7.00. The first thing to do is to calculate where the equivalence point is.The following will do that.
mL acid x M acid = mL base x M base
mL acid x 0.125 = 30.00 x 0.150
mL acid = ?

Then set up an ICE chart for each addition of acid to the base. We start with how many moles of KOH? That is M x L = 0.150 x 0.0300 = .0045
Additions of HClO4 are
30.00 x 0.125 = 0.00375
35.00 --- you can fill
36.00 --- you can fill
40.00 --- you can fill.

..........KOH + HClO4 ==>KClO4 + H2O
initial..0.0045..0.........0......0
add 30..........0.00375............
change.-0.00375 -0.00375..+0.00375..etc
equil....0.00075..0...0.00375..0.00375

The ICE chart tells you what you have at this point in the titration which for 30.00 mL is a solution of KOH and KClO4 in a volume of 60.00 mL (30.00 acid _ 30.00 base).
M KOH = moles/L = 0.00075/0.0600 = 0.0125M KOH. Then pOH = ? and obtain pH from pH + pOH = pKw = 14.
Everything up to equilvanece point is done this way. The equivalence point, as I pointed out above has a pH of 7.00, and everything after the equivalence point done the same way EXCEPT you should note that for those points its the HClO4 in excess. Post your work if you get stuck.

• Chemistry - ,

I am doing similar homeowrk and I was wondering how you calculate the ph after the equivalence point?

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