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Chemistry 135
Clark College
YELLOW Exam 1 SOLUTION
Please work in the spaces provided, or on the backs of pages. You MUST show your work
for full credit! Put your answers in the boxes provided.
1. Natalie O’Henry was using KHP to standardize a solution of Ba(OH)2. After 4 titrations,
she calculated the following 4 normality values for her titrations: 0.8719 N, 0.8653 N, 0.8677
N, and 0.8794 N. Round her values to ppth precision, and then determine the mean,
standard deviation, and RSD, retaining all values.
Note: If you are using a non-programmable calculator that handles statistics functions, you may use
your calculator to perform the calculations for this problem. Please state that you are using the
calculator, and show the set up of the equations/formula.
x
0.865 N
0.868 N
0.872 N
0.879 N
xbar = 0.871
Standard Deviation =
x – xbar
-0.006
0.003
0.001
0.008
! (x - x )
2
(x-xbar)2
3.6 x 10-5
9 x 10-6
1 x 10-6
6.4 x 10-5
Σ = 1.1 x 10-4
! ( 1.1 x 10 )
-4 2
=
= 0.00606
n-1
4-1
Rounded to the same number of decimal places as the mean, SD = 0.006 N
rounded SD
0.006
x 1000 =
x 1000 = 6.88
mean
0.871
Rounded to the same number of significant digits as SD, RSD = 7 ppth
RSD =
N Ba(OH)2
0.865 N
0.868 N
Mean =
0.871 N
Std. Deviation =
0.006 N
RSD =
7 ppth
0.872 N
0.879 N
2. What was the average molarity of Natalie’s Ba(OH)2 solution from problem #1?
0.871 eq 1 mol Ba(OH)2
x
= 0.436 M
L
2 eq OH M Ba(OH)2 = 0.436 M
Exam 1
Spring 2007
Page 1 of 2
Chemistry 135
Clark College
3. A 0.6834 g sample of KHP (99.97% pure) required 31.93 mL of NaOH to titrate to a pale
pink color. Determine the normality of NaOH used for this titration, and report your value
to ppth precision. (Eq. wt. KHP = 204.23 g/eq)
0.6834 g sample x
99.97 g KHP
1 eq KHP
1 eq NaOH
1
x
x
x
= 0.1048 N NaOH
100 g sample 204.23 g KHP
1 eq KHP
0.03193 L
N NaOH = 0.1048 N
4. Is the standard deviation a measure of accuracy or precision in your data?
Precision
The standard deviation is a measure of how close together your results are – how narrow or wide the
values are distributed. This measures the precision of your data. It says nothing about how close your
data is to an “accepted” value – that is the accuracy!
5. Which of the following is not an error when using an analytical balance to weigh out your
KHP?
a. Not closing the balance “doors”.
b. Waiting for the mass to stop changing.
c. Spilling solid on the balance pan.
d. Incomplete transfer of KHP from the paper to the Erlenmeyer flask.
Credit was given for both answers. You should wait for the mass to STOP changing before recording a
mass. Incomplete transfer would be an error, but it is technically not an error when using an analytical
balance. It occurs after using the balance!
6. You are calibrating a 100-mL volumetric flask, and come up with the following 5 volumes
(in mL): 99.7, 99.9, 100.0, 100.1, 99.1. Perform a Q-test and the 5-ppth test to determine if
the last value can be discarded. Then, calculate the mean volume from the appropriate
data. Q90 = 0.642.
Mean = 99.8 mL
Q=
value - neighbor
99.1 - 99.7
=
= 0.6 > Q 90 Passes!
range
99.1 - 100.1
5 ppth test:
Exam 1
value - mean
99.1 - 99.8
x 1000 =
x 1000 = 7 ppth Fails!
mean
99.8
Qcalc =
0.6
Does it pass? (Y/N)
Yes
ppth test value =
7 ppth
Does it pass? (Y/N)
No
Mean =
99.8 mL
Spring 2007
Page 2 of 2