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Bad White [126]
3 years ago
10

I am having some trouble figuring out how to approach the following problem: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is ti

trated with 0.500 M HNO3. Calculate the pH after the addition of 17.0mL of HNO3.
Chemistry
1 answer:
Tanya [424]3 years ago
4 0
<span>We look at the end of the day:

n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>
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3 years ago
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vodomira [7]

Answer:

Wet deposition

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When they reach the soil, eventually, the water will  evaporate or will be absorbed by soil, so the chemicals that are left behind, will reach the surface of Earth.

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3 years ago
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What is the pH of a sample with [OH-] = 2.5 x 10-11 M?
sveticcg [70]

Answer:

The pH of the sample is 3,4.

Explanation:

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3 years ago
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