Question

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 acidity 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

Answer 1

The concentration of sulfuric acid is 0.0033 M

Step-by-step 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