Answer:
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
Explanation:
Volume of NaOH = 1.7 ml = 0.0017 L
Molarity of NaOH = 0.0811 M
Moles of NaOH = n
[tex]0.0811 M=\frac{n}{0.0017 L}[/tex]
n = 0.0001378 mol
[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]
According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.
Then 0.0001378 mol of NaOH will neutralize:
[tex]\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol[/tex] of sulfuric acid.
Concentration of sulfuric acid in the acid rain sample: x
[tex]x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L[/tex]
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
