Environmental studies usually involve an analysis of precipitation and its response to pollution. To quantify the degree of contamination in natural rain water or snow, titration is used. The process is quick and results are reliable. Since most titration processes do not require expensive or specialized equipment, the test can be performed often and in different areas with relatively little effort. a. A 1000.0 mL sample of lake water is titrated using 0.100 mL of a 0.100 M base solution. What is the molarity of the acid in the lake water? Show all work. b. Based on the molarity of the acid calculated above, what is the pH of the lake water?

a. Molarity =0.00001 M

b. pH of lake water =5

Given:

Molarity of base solution = 0.1 M

Volume of the base solution = 0.1 ml or 0.0001 litre

Volume of lake water = 1000ml  or 1 litre

Molarity of the lake water, = ?

a. Using the formula for titration:

M₁V₁=M₂V₂

So,

The molarity of lake is 0.00001M.

b. The pH of the lake water will be calculated by using the following formula:

The pH of the lake water is 5.

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The molarity of the lake water is 0.00001 M and the pH of lake water is 5.

The lake water is acidic.

Explanation:

Data given:

molarity of base solution Mbase = 0.1 M

volume of the base solution Vbase = 0.1 ml or 0.0001 litre

volume of lake water Vlake = 1000ml  or 1 litre

molarity of the lake water, Mlake = ?

Using the formula for titration:

Mbase X Vbase = Mlake X

Mlake =

Putting the values in the equation:

Mlake =

Mlake = 0.00001 M

The pH of the lake water will be calculated by using the following formula:

pH = - []

pH = - [ 0.00001]

pH = 5

14 days is the half-life of phosphorus-32

elements make up everything, and they combine to make compounds.

the answer is c