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elena-14-01-66 [18.8K]
3 years ago
6

Titration of a 24.0 mL sample of acid rain required 1.8 mL of 0.0882 M NaOH to reach the end point. If we assume that the acidit

y of the rain is due to the presence of sulfuric acid, what was the concentration (in M) of sulfuric acid in this sample of rain
Chemistry
1 answer:
wolverine [178]3 years ago
3 0

Answer:

The concentration of sulfuric acid is 0.0033 M

Explanation:

Step 1: Data given

Volume of the sulfuric acid (H2SO4) = 24.0 mL = 0.024 L

Volume of NaOH = 1.8 mL = 0.0018 L

Molarity of NaOH = 0.0882 M

Step 2: The balanced equation

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Step 3: Calculate the concentration of H2SO4

b*Ca*Va = a * Cb * Vb

⇒b = the coefficient of NaOH = 2

⇒Ca = the concentration of H2SO4 = TO BE DETERMINED

⇒Va = the volume of H2SO4 = 0.024 L

⇒a = the coefficient of H2SO4 = 1

⇒Cb = the concentration of NaOH = 0.0882 M

⇒Vb = the volume of NaOH = 0.0018 L

2* Ca * 0.024 = 1 * 0.0882 * 0.0018

0.048 * Ca =  0.00015876‬

Ca = 0.0033 M

The concentration of sulfuric acid is 0.0033 M

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Explanation:

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