How many mL of 0.5 M sodium hydroxide would it take to react completely with 50 mL of 0.3 M sulfuric acid?
To answer this question, we need to determine the balanced chemical equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4).
The balanced equation is:
2 NaOH + H2SO4 -> Na2SO4 + 2 H2O
From the balanced equation, we can see that for every 2 moles of NaOH, we need 1 mole of H2SO4.
We can use the equation:
moles = concentration × volume (in liters)
to calculate the number of moles of H2SO4:
moles of H2SO4 = 0.3 M × 0.050 L = 0.015 moles
Since the stoichiometry of the reaction is 2:1 (NaOH:H2SO4), we need 2 times the number of moles of NaOH:
moles of NaOH = 2 × moles of H2SO4 = 2 × 0.015 moles = 0.030 moles
Now we can use the equation:
volume (in liters) = moles / concentration
to calculate the volume of 0.5 M NaOH needed:
volume of NaOH = 0.030 moles / 0.500 M = 0.060 L = 60 mL
Therefore, it would take 60 mL of 0.5 M sodium hydroxide to react completely with 50 mL of 0.3 M sulfuric acid.