Respuesta :
Let's find the number of moles of 4.14 g of H2SO4 using its molar mass:
[tex]4.14gH_2SO_4\cdot\frac{1molH_2SO_4}{98.09\text{ }gH_2SO_4}=0.042\text{ mol }H_2SO_4\text{.}[/tex]With this value, we can calculate how many moles of NaOH (base) we have to calculate its molarity. In the chemical reaction, you can see that 2 moles of NaOH react with 1 mol of H2SO4, so doing a rule of three:
[tex]\begin{gathered} 2\text{ moles NaOH}\to1\text{ mol }H_2SO_4 \\ \text{? moles NaOH}\to0.042\text{ moles }H_2SO_4\text{.} \end{gathered}[/tex]And the calculation would be:
[tex]0.04\text{2 moles }H_2SO_4\cdot\frac{2\text{ moles NaOH}}{1\text{ mol }H_2SO_4}=0.08\text{4 moles NaOH.}[/tex]Now we can calculate the molarity using the formula:
[tex]\text{Molarity}=\frac{moles\text{ of solute}}{\text{Volume in liters}}.[/tex]But before doing that, we need to convert 85 mL to liters. Remember that 1 liter equals 1000 mL:
[tex]85.00mL\cdot\frac{1\text{ L}}{1000\text{ mL}}=0.085\text{ L.}[/tex]And finally, we can calculate the molarity like this:
[tex]\begin{gathered} \text{Molarity}=\frac{0.084\text{ moles}}{0.085\text{ L}}, \\ \text{Molarity}=0.988\text{ M.} \end{gathered}[/tex]The answer is that the molarity of the base before neutralizing it is 0.988 M.
