a. The pH meter was improperly calibrated, so that the pH readings on the instru
ID: 500163 • Letter: A
Question
a. The pH meter was improperly calibrated, so that the pH readings on the instrument were 0.3 pH units too low.
A. pKa would be too low and Ka would be too high
B. pKa would be too high and Ka would be too low
C. neither would change
b. An endpoint/equivalence point was erroneously read as 27.20 mL whereas the reading was actually 26.80 mL.
A. pKa would be too low and Ka would be too high
B. pKa would be too high and Ka would be too low
C. neither would change
D. none of these
c. In Step 2, instead of adding 75 mL of water, 150 mL of water was added.
A. pKa would be too low and Ka would be too high
B. pKa would be too high and Ka would be too low
C. neither would change
D. none of these
Please explain why you chose the answer you chose, thank you!
Explanation / Answer
a) effect of Ka with a pH is marking 0.3 lower...
example:
pH = pKa at half equivalence point...
if real pKa was 4.0, then pH = 4.0
Bu since it is marking0.3 lower, then at the same volume, you will expect a pH = 4-0.3 = 3.7
so the pKa = 3.7, which is false, since it is LOWER
meaning that
Ka = 10^-pKa =10^-3.7 will be HIGHER
choose a
b)
Actual V = 26.8 mL
V false = 27.20 mL
if this is the case, then
the half equivalence point will be lager than normal
will have a higher pH
so
pK = pH ( higher )
meaning that
pKa is highly calculated
so
Ka =10^-pKa = will be LOWE, choose b
C)
We require more data t what Step 2 is... but:
if this is for the base --> no effect
if this is for the acid --> acid gets diluted, therefore accounted for a weaker "acid" expect lower Ka
for both mixtures --> o effect, since both volumes cance each other in Macid*Vacid = Mbase*Vbase
since Vacid = Vbase
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