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AP Chemistry 2015 Titration Problem
Transcript of AP Chemistry 2015 Titration Problem
Sydney Pham & Rachel Tindall
(f) The pH of the soft drink is 3.37 after the addition of the (aq). Which species, or , has a higher concentration in the soft drink? Justify your answer.
(b) A total of 29.95 mL of 1.25 M HCl(aq) is required to reach the equivalence point. Calculate [ ] in the stock solution.
(e) The initial pH and the equivalence point are plotted on the graph below. Accurately sketch the titration curve on the graph below. Mark the position of the half-equivalence point on the curve with an X.
(d) Calculate the pH at the half-equivalence point.
(Question 3 on Free Response)
Potassium sorbate, (molar mass 150. g/mol) is commonly added to diet soft drinks as a preservative. A stock solution of (aq) of known concentration must be prepared. A student titrates 45.00 mL of the stock solution with 1.25 M HCl(aq) using both an indicator and a pH meter. The value of Ka forsorbic acid, , is 1.7 × 10−5.
AP Question #3
(a) Write the net-ionic equation for the reaction between (aq) and HCl(aq).
29.95 mL HCl 1L HCl 1.25 mol HCl 1 mol potassium sorbate = 0.0374 mol
1000 mL HCl 1L HCl 1 mol HCl potassium sorbate
(c) The pH at the equivalence point of the titration is measured to be 2.54 Which of the following indicators would be the best choice for determining the end point of the titration? Justify your answer.
Bromothymol blue 7.0
Methyl red 5.0
Thymol blue 2.0
Methyl violet 0.80
Thymol blue would be the most appropriate indicator to use for this titration because it is the indicator that has the endpoint (pKa of 2) closest to the equivalence point (pH of 2.54) of the titration.
pH = pKa
pH = -log(1.7 x 10 ) = 4.77
pH = pKa + log [base]
because the pKa is 4.77, the second part of the above equation must be negative in order to produce the pH of 3.37, requiring the ratio of the salt to acid to be less than 1, requiring the acid to have a higher concentration.