How many milliliters of 2.3 M NaOH are needed to neutralize 4.2 mL of a 5.4 M H2SO4 solution?
To solve this problem, we can use the concept of stoichiometry and balanced chemical equations. The balanced equation for the neutralization reaction between NaOH and H2SO4 is:
2 NaOH + H2SO4 -> Na2SO4 + 2 H2O
From the balanced equation, we can see that the ratio of NaOH to H2SO4 is 2:1. This means that for every 2 moles of NaOH, we need 1 mole of H2SO4 to neutralize the solution.
First, we need to calculate the number of moles of H2SO4 in the 4.2 mL of 5.4 M H2SO4 solution:
moles of H2SO4 = volume (L) x concentration (M)
= 4.2 mL / 1000 mL/L x 5.4 mol/L
= 0.02268 mol
Since the ratio of NaOH to H2SO4 is 2:1, we need half as many moles of NaOH as H2SO4 to neutralize the solution:
moles of NaOH = 0.02268 mol / 2
= 0.01134 mol
Now, we can calculate the volume of 2.3 M NaOH solution needed to supply 0.01134 moles of NaOH:
volume (L) = moles / concentration (M)
= 0.01134 mol / 2.3 mol/L
= 0.00493 L
But the volume is given in milliliters (mL), so we convert the volume to mL:
volume (mL) = 0.00493 L x 1000 mL/L
= 4.93 mL
Therefore, you would need approximately 4.93 mL of 2.3 M NaOH to neutralize 4.2 mL of a 5.4 M H2SO4 solution.