What is the concentration(in Molarity)

Question

  H₂SO₄ + 2 Na(OH)      2H₂O + Na₂SO₄

Based on the above equation, calculate the concentration (in M) of a 49.1 mL solution of Na(OH) if 50.0 mL of 2.95 M H₂SO₄ is required to react completely.

Answer 1

The balanced equation is H2SO4 + 2NaOH → Na2SO4 + H2O. For every mole of H2SO4 reacted, 2 moles of NaOH are required for a complete reaction.

If we calculate the number of moles of H2SO4 used, then we can calculate the number of moles of NaOH that were present. From that and the volume of NaOH initially present, we can determine its molarity.

Moles of H2SO4 needed = 2.95 mol/L H2SO4 / 1000 mL/L x 50 mL = 0.148 moles H2SO4.

According to the balanced equation, we need twice as many moles of NaOH for complete reaction, so there are 2 x 0.148 moles NaOH in the 49.1 mL of NaOH solution. The number of moles NaOH per liter (molarity) is then 2 x 0.148 moles / 49.1 mL x 1000 mL/L = 6.03 M.

Always check math.

Answer 2

This answer given is mostly correct, I just wanted to point out that there is an accidentally dropped 2 before the H2O (it doesn't have an effect on the math of this problem, but it is important). Also that if you don't round the moles of H2SO4 until the end, that the molarity of Na(OH) required is 6.01 M. It's not a big change, but it could account for a difference in answers if your answer doesn't match.

My work is as follows (it's the same):

2.95 mol H2SO4 1 L 50 mL 2 mol Na(OH)

----------------------- X ---------- X -------- X ------------------- = 0.295 mol Na(OH)

L 1000mL 1 mol H2SO4

2.95 mol Na(OH) 1000 mL

----------------------- X ---------- = 6.01 Na(OH)

49.1mL Na (OH) 1L

Mr. Anthony explained how to do the problem very well, I was just hoping that reformatting might make the concepts easier to see.