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Chapter 4: Q79E (page 224)

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811MNaOH 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 of sulfuric acid in this sample of rain?

Short Answer

Expert verified

0.0068 M was the concentration of sulfuric acid in the sample of rain.

Step by step solution

01

Given Data

\({M_a}{V_a} = {M_b}{V_b}\)

a = acid = H2SO4 and b = base=NaOH

Volume of acid = 20 mL

Volume of base = 1.7 mL

Molarity of base = 0.0811 M

Molarity of acid = ?

02

Determine the concentration of sulfuric acid

\(\begin{aligned}{}{M_a}{V_a} &= {M_b}{V_b}\\{M_a} \times 20\,mL &= 0.0811\,M \times 1.7\,mL\\{M_a} &= \frac{{0.0811\,M \times 1.7\,mL}}{{20\,mL}}\\{M_a} &= 0.0068\end{aligned}\)

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