Question

1. For the following replicate measurements, 2.4, 2.1, 2.1, 2.3, and 1.5, the mean is 2.08 and the standard deviation is 0.349. Develop a 90% confidence interval using the t-distribution to encase the true value. Compare this to a 90% confidence interval developed using a normal distribution.

Values of t for Various Levels of Probability

Degrees of Freedom	80%	90%	99%	99.9%
1	3.08	6.31	63.7	637
2	1.89	2.92	9.92	31.6
3	1.64	2.35	5.84	12.9
4	1.53	2.13	4.60	8.6
5	1.48	2.02	4.03	6.86
6	1.44	1.94	3.71	5.96
7	1.42	1.90	3.50	5.40
8	1.40	1.86	3.36	5.04
9	1.38	1.83	3.25	4.78

2. Consider the standardization of NaOH by KHP (MW = 204.22g). 13.3ml, 13.5ml, and

13.0ml of NaOH were neutralized by 0.716, 0.739, and 0.711 grams of KHP. What is the

concentration of the NaOH solution? Express the uncertainty in your answer using both

propagation of error (assume uncertainty in the volume of solution dispensed by the burette

is 0.1 ml) and statistics (95% confidence interval)

Answer 1

Ques 1:

Since the number of observations (n) = 5, the degree of freedom = n-1 = 4

For degree of freedom = 4, the t-value for 90% probability = 2.13

Now, the 90% confidence interval to encase the true value = Mean ± t X standard deviation/√n = 2.08 ± 2.13 X 0.349/√5 = 2.08 ± 0.33

For a normal distribution with 90% probability, t = 1.645

90% confidence interval developed using a normal distribution = 2.08 ± 1.645 X 0.349/√5 = 2.08 ± 0.25

Therefore, the 90% confidence interval developed using normal distribution is narrower.

Ques 2:

Using the molarity equation: M₁V₁ = M₂V₂

Mean value of NaOH standardization V₁ = (13.3 + 13.5 + 13)/3 = 13.26 mL

Mean value of KHP = (0.716 + 0.739 + 0.711)/3 = 0.722 g

Therefore average number of moles of KHP = 0.722/MW = 0.722/204.22 = 0.0035

Now, to solve further, volume of KHP solution is required. If you have that volume, then put it in the equation: M₁V₁ = M₂V₂ and calculate M₁ which is the concentration of NaOH solution.

Now, M₁ ± 0.1 is the required error.