How many milliliters of 2.00 M NaHO solution are needed to neutralize a 8.00 mL of 3.50 M H2SO4 solution?
H2SO4 + 2NaOH --> 2H2O + Na2SO4
mols H2SO4 = M x L = ?
Using the coefficients in the balanced equation, convert mols H2SO4 to mols NaOH.
Then M NaOH = mols NaOH/L NaOH. You know mols and M, solve for L and convert to mL.
To find the volume of NaHO solution needed to neutralize the H2SO4 solution, we can use the concept of the balanced chemical equation between NaOH and H2SO4. The balanced equation is:
2NaOH + H2SO4 → Na2SO4 + 2H2O
From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of H2SO4. Therefore, the mole ratio (stoichiometry) is 2:1.
First, let's find the moles of H2SO4 present in the given solution:
Moles of H2SO4 = Molarity × Volume
= 3.50 M × 8.00 mL (converting mL to L)
= 0.028 moles
Since the stoichiometry between NaOH and H2SO4 is 2:1, we know that 2 moles of NaOH is required to react with 1 mole of H2SO4.
Now, let's find the volume of NaOH solution needed to neutralize the H2SO4 solution:
Volume of NaOH solution = (Moles of H2SO4 ÷ Molarity of NaOH) × Volume unit conversion
Substituting the values into the equation:
Volume of NaOH solution = (0.028 moles ÷ 2.00 M) × 1000 mL (converting L to mL)
= 14 mL
Therefore, 14 mL of 2.00 M NaOH solution is needed to neutralize 8.00 mL of 3.50 M H2SO4 solution.