How many mL of 0.5 M sodium hydroxide would it take to react completely with 50 mL of 0.3 M sulfuric acid?
The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4) is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
From the balanced equation, we can see that the ratio of NaOH to H2SO4 is 2:1. This means that 2 moles of NaOH reacts with 1 mole of H2SO4.
To calculate the moles of H2SO4 in 50 mL of 0.3 M sulfuric acid, we use the equation:
Moles = Volume (L) x Concentration (M)
Moles of H2SO4 = 0.050 L x 0.3 M
Moles of H2SO4 = 0.015
Since the ratio of NaOH to H2SO4 is 2:1, we need to have half the number of moles of NaOH compared to H2SO4.
Moles of NaOH = 0.015 / 2
Moles of NaOH = 0.0075
Now, let's calculate the volume of 0.5 M NaOH needed to react completely with the H2SO4.
Volume (L) = Moles / Concentration (M)
Volume of NaOH = 0.0075 / 0.5
Volume of NaOH = 0.015 L
To convert the volume into milliliters (mL), we multiply by 1000.
Volume of NaOH = 0.015 L x 1000 mL/L
Volume of NaOH = 15 mL
Therefore, it would take 15 mL of 0.5 M NaOH to react completely with 50 mL of 0.3 M H2SO4.