How many mL of 1.67M sulfuric acid would it take to neutralize 26.6mL of 3.12M sodium hydroxide?

2NaOH + H2SO4 ==> 2H2O + Na2SO4

mols NaOH = M x L = ?
Use the coefficients in the balanced equation to convert mols NaOH to mols H2SO4. (You can see mols H2SO4 = 1/2 mols NaOH)
Then M H2SO4 = mols H2SO4/L H2SO4. You know M and mols, solve for L and convert to mL.

To find the answer, we can use the concept of molarity (M) and the equation for neutralization reactions.

First, let's understand the balanced equation for the neutralization reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH):

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

From the balanced equation, we can see that one mole of sulfuric acid reacts with two moles of sodium hydroxide.

Given information:
- Volume of sodium hydroxide (NaOH) = 26.6 mL
- Molarity of sodium hydroxide (NaOH) = 3.12 M
- Molarity of sulfuric acid (H₂SO₄) = 1.67 M

To find the number of moles of sodium hydroxide, we can use the following formula:

moles = Molarity × Volume (in liters)

Converting the given volume of sodium hydroxide to liters:

Volume of NaOH = 26.6 mL = 26.6 mL × (1 L / 1000 mL) = 0.0266 L

Calculating the number of moles of sodium hydroxide:

moles of NaOH = 3.12 M × 0.0266 L = 0.082992 moles

Since the ratio of sodium hydroxide to sulfuric acid is 1:1, we know that the number of moles of sulfuric acid required to neutralize the sodium hydroxide is also 0.082992 moles.

Finally, we need to calculate the volume of sulfuric acid using the molarity of sulfuric acid:

Volume of H₂SO₄ = moles / Molarity
Volume of H₂SO₄ = 0.082992 moles / 1.67 M

Using a calculator, we get:
Volume of H₂SO₄ ≈ 0.0498 L = 49.8 mL

Therefore, it would take approximately 49.8 mL of 1.67 M sulfuric acid to neutralize 26.6 mL of 3.12 M sodium hydroxide.