First part Second part General Chemistry II Workshop 5: Titration Curves A pH or
ID: 490680 • Letter: F
Question
First partSecond part General Chemistry II Workshop 5: Titration Curves A pH or titration curve can be produced by plotting the pH of a solution (containing an analyte) versus the volume of added titrant. The equivalence or end point of a titration occurs when enough titrant has been added to consume all the acid or base initially present in the analyte. You know you have reached the endpoint when you see a significant change in the pH ofthe solution Part 1 Let's begin by analyzing a titration curve (2 pts) When your analyte is a weak acid or base the titration curve shows the effects of buffering before the equivalence point. The midpoint of the titration occurs in the buffer zone and is the point at which equal to the pKa of the analyte. In the case of a weak acid-strong base titration the pH at the equivalence point will always be higher than 7. This is because of the basic properties of the a (the conjugate base of the weak acid) that is present in the solution at this point. To the right is a titration curve for the Titration of unknown acid titration of 25.0 mL of unknown compound X with 0.200 M NaoH. Use this titration curve to answer the following questions 10 (a) What is the identity of compound X? pH HINT: Use the Chempendix on Sapling to look at the Ka values for possible weak acids. (1 pt) 10 20 30 40 50 G,3 y lo volume of 0.200 M NaoH added, mL hkah AciD (b) What is the starting concentration of compound X? (1 pt) At equ valent nt (moles of Nou. (molas of co H sco D) oo, 6 m les NaOH m Before you move on to part 2 please make sure EVERYONE in your group is comfortable CH3 predicting the identity and starting concentration of an analyte from the midpoint titration curve. This is a very important skill! You will need the correct complete part 2 of the workshop. What can you do to make sure you have th concentration to correct answer?
Explanation / Answer
0 mL
for 15 mL --> A- = HA
30.75 mL --> equivalence
40 mL --> excess base
so:
at 0 mL, there is only acid, HA, and some H+; A- in solution, so pH active are those of H+ only... pH from graph = 2
at V = 20 mL
there are
HA, H+ and A- mainly.... so pH active species are both, Ha and A- since this is a buffer
so
pH = 4.5 approx
at V = equivalence point
the major species rpesent:
A- and H2O... also HA due to hydrolysis in some extent and some OH- due to the same process
pH active is only OH- ions... pH = 8.5 approx
then
finally
V = 40 mL , exces base is present 12.5 approx
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