# Titration

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Consider the titration of 100.0 mL of 0.100 M NaOH with 1.00 M HBr. Find the pH at the following volumes of acid added and make a graph of pH versus Va: Va = 0, 1, 5, 9, 9.9, 10, 10.1, and 12 mL.

• Titration -

This is the titration of a strong base with a strong acid.
Begin: 100 mL x 0.100M NaOH = 10 millimoles.
millimoles HBr added = 1.00M x mL = ?

.............NaOH + HBr ==> NaBr + H2O
initial......10mmol...0.......0......0
added.................0
change........-0............
equil.........10mmol..........

The trick here is to know what is in the solution when you are ready to determine the pH. In the case of V HBr = 0, you can see we have a solution of 10 mmoles NaOH in 100 mL = and M = mmol/mL = 10/100 = 0.1 M NaOH.
That will give you (OH^-) = 0.1 which is pOH of 1 and pH of 13.

Next point.
............NaOH + HBr ==> NaBr + H2O
initial......10.....0.......0......0
added.............1x1=1...........
change.......-1....-1.......+1.....+1
equil........9.......0.......1......1
So you have a solution of 9 mmoles NaOH and 1 mmole NaBr. NaBr is not hydrolyzed (either Na or Br) so M NaOH = mmol/mL = 9/101 = 0.0891; pOH = -log(OH^-) = 1.05 and pH + pOH = pKw = 14 so pH = 12.94 (the 101 for mL total volume is 100 to start with NaOH + 1 mL added from HBr.)

That is how you do all of the points from the beginning to the equivalence point.
The equivalence point is determined by the hydrolysis of the salt. Since it isn't hydrolyzed, the pH at the equivalence point is just the pH of pure water which is pH = 7.0.

All points past the equivalence point are done the same way with the only difference being that all of the NaOH is used and you are adding successively larger excesses of HBr.You can find which point is the equivalence point by mLNaOH x MNaOH = mLHBr x MHBr and solve for mL HBr.
Post your work if you need additional help.

• Titration -

1294

• Titration -

pH=12.94

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