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calculate the pH of a solution prepared by dissolving 1.75g of acetate, CH3COONa, in 60.5mL of 0.25M acetic acid, CH3COOH(aq). A
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Answer:
a)
b) Lower endpoint: 0.422cc/m³
Upper endpoint: 0.452 cc/m³
Step-by-step explanation:
Population is approximately normal, so we can find the normal confidence interval.
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
. This is the critical value, the answer for a).
Now, find M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 0.437 - 0.015 = 0.422cc/m³.
The upper end of the interval is the sample mean added to M. So it is 0.437 + 0.015 = 0.452 cc/m³.
b)
Lower endpoint: 0.422cc/m³
Upper endpoint: 0.452 cc/m³