The first order of business is to determine exactly where you are on the titration curve after 18.0 mL NaOH. A starting point is to determine exactly where the first and second equivalence points are along the titration curve.
How much acid do we have? M x L = mols = 0.360 x 0.030 = 0.0108 moles acid.
So the first eq point will come at 0.0108 moles of NaOH and the second one will come after another 0.0109 moles NaOH.
moles NaOH = 0.0108 = M x L. M = 0.280. Solve for L and I get something like 38.57 mL. Therefore, we know that 18.0 mL is between the starting point and the first eq point. The equation we are working with is something like this.
H2A + NaOH ==> NaHA + H2O
Yes, now use the ICE table to determine moles start, moles NaOH added (at 18.0 mL), moles NaHA formed, moles H2A remaining. Since this is before the first eq point (if I've not done something wrong), you will have formed some of the salt and you will have some of the acid remaining. Aha! a buffer. Use the Henderson-Hasselbalch equation to solve for the pH.
Question: Consider the titration of 30.00 ml of .360 M. H2C6H6O6 (abscorbic acid; K1= 6.8e-5; K2=2.8E-12) solution with .280M NaOH. Note the weak acid, H2C6H6O6, is being titrated with the strong base, sodium hydroxide. the neutralization reaction are: H2C6H6O6 + NaOH goes to NaHC6O6 + H20 and NaHC6H6O6 + NaOH goes to Na2C6H6O6 + H2O. A) Compute the pH after 18.0 ml. of NaOH have been added.
Not sure what numbers go where and how to do the problem? Do you need an ICE table?
2 answers
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1 answer