Write the balanced chemical equation for the reaction as follows:
[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O[/tex]Given:
Concentration of sodium hydroxide =1.50 mol/L
Volume of sodium hydroxide = 20 mL
Concentration of sulfuric acid = x
Volume of sulfuric acid = 20 mL
Firstly, we will determine the moles that is in 20 mL of 1.5 M NaOH:
[tex]\begin{gathered} 1.5\text{ }mole=1000mL \\ x\text{ }mol=20mL \\ \\ x=\frac{1.5mol\times20mL}{1000mL} \\ \\ x=0.03mol\text{ }NaOH \end{gathered}[/tex]Based on the balanced chemical equation and the moles of NaOH we can use stoichiometry (ratios) to determine the moles of 20 mL sulfuric acid:
[tex]0.03mol\text{ }NaOH\times\frac{1mol\text{ }H_2SO_4}{2mol\text{ }NaOH}=0.015\text{ }mol[/tex]Now that we know how many moles are in 20 mL of H2SO4, we can determine the moles in 1000mL (initial concentration):
[tex]\begin{gathered} 0.015\text{ }mol=20\text{ }mL \\ x\text{ }mol=1000\text{ }mL \\ x=\frac{0.015mol\times1000mL}{20mL} \\ \\ x=0.75mol•L^{-1} \end{gathered}[/tex]Answer: D) 0.75 mol/L,
