1. A) Assume Class A glassware for all calculations (tolerance of +/- 0.1%). Jan
ID: 986196 • Letter: 1
Question
1. A) Assume Class A glassware for all calculations (tolerance of +/- 0.1%). Janice prepares 500 mL of a 45 millimolar stock solution of HCl by diluting 50.0 mL of a 450 millimolar stock up to 500.0 mL in a volumetric flask. Solve for pH of the prepared solution with associated error based on these dilutions.
B) Christine takes Janice's sample from Problem A and measures its pH on a pH meter that reports with an intrinsic RSD of 2.3%. Taking that into account with the variability reported in Problem A, estimate the pH and relative standard deviation Christine will report for this sample.
C) We'll assume the pH and error generated in problem 4 were the result of 10 replicate measurements. The solution is too acidic for Christine to use if the pH is below 4.75. Is the pH estimated from problem 4 less than pH 4.75 at the 95% level of confidence? Show your work, including your hypothesis statements.
Explanation / Answer
The initial concentration of HCl = 450
Volume measured = 50mL +/- 0.1%
Volume make up = 500mL +/- 0.1%
M1V1 = M2V2
M1 = M2XV2 / V1
So we have the error propagation due to multiplication and division of the given volume
error = uconcentration / concentration = (0.1 / 50)2 + (0.1 / 500)2
error = (0.000004 + 0.00000004 )^1/2 = 0.002 %
pH = -log[H+]
[H+] = 45millimolar = 45 X10^-3 molar
pH = -log [45X10^-3] = 1.34
so pH error = 0.002% of 1.34 = 0.0000268
B)
RSD = 100 X standard deviation / average value
We need different pH values for calcualtion of RSD
However if RSD = 2.3%
So RSD = 2.3 X 1.34 / 100 = 0.0308