How many mL of 0.40M NaOH are need to neutralize 30.0 mL of 0.50 M H2SO4
according to the following reaction.
2NaOH(aq) + H2SO4 (aq)----Na2SO4 (aq) + 2H2O(l)
mols H2SO4 = M x L = ?
mols NaOH = twice that (from the coefficients)
M NaOH = mols NaOH/L NaOH. You know M and mols; solve for L and convert to mL. Click post answer
0.0048 ml
To find out how many mL of 0.40 M NaOH are needed to neutralize 30.0 mL of 0.50 M H2SO4, we can use the concept of stoichiometry which relates the number of moles of two substances in a chemical reaction.
First, let's calculate the number of moles of H2SO4 present in 30.0 mL of 0.50 M H2SO4. We can use the formula:
moles = concentration × volume
moles of H2SO4 = 0.50 M × 0.0300 L
moles of H2SO4 = 0.015 moles
According to the balanced equation, the stoichiometric ratio of NaOH to H2SO4 is 2:1. So, for every 2 moles of NaOH, we need 1 mole of H2SO4.
Therefore, we need half the number of moles of NaOH compared to the number of moles of H2SO4, as they are in a 1:2 ratio (according to the balanced equation).
moles of NaOH = 0.015 moles / 2
moles of NaOH = 0.0075 moles
Finally, we can use the formula for concentration to calculate the volume of 0.40 M NaOH required:
volume = moles / concentration
volume = 0.0075 moles / 0.40 M
volume = 0.01875 L
To convert this volume into mL, we can multiply by 1000:
volume = 0.01875 L × 1000
volume = 18.75 mL
Therefore, 18.75 mL of 0.40 M NaOH is needed to neutralize 30.0 mL of 0.50 M H2SO4.