What is the concentration of an acetic acid solution if 11.2 mL of 1.002 Molarity sodium hydroxide solution is needed to completely neutralize 5.1 mL of the acid?

To determine the concentration of the acetic acid solution, we can use the concept of stoichiometry.

The balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is as follows:

CH3COOH + NaOH -> CH3COONa + H2O

From the balanced equation, we can see that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide.

First, let's calculate the number of moles of sodium hydroxide used:

Moles of sodium hydroxide = Molarity × Volume
= 1.002 mol/L × 11.2 mL
= 0.0112 L × 1.002 mol/L
≈ 0.0112 mol

Since the mole ratio between acetic acid and sodium hydroxide is 1:1, the number of moles of acetic acid is also 0.0112 mol.

Now, let's calculate the concentration of the acetic acid solution:

Concentration = Moles/Volume
= 0.0112 mol / 5.1 mL
= 0.0112 mol / 0.0051 L
≈ 2.2 M

Therefore, the concentration of the acetic acid solution is approximately 2.2 M.

To determine the concentration of the acetic acid solution, you need to use the concept of stoichiometry and the balanced equation of the neutralization reaction between acetic acid and sodium hydroxide.

The balanced equation for the neutralization reaction is:

CH3COOH (acetic acid) + NaOH (sodium hydroxide) -> CH3COONa (sodium acetate) + H2O (water)

From the balanced equation, you can see that the ratio of moles of acetic acid to moles of sodium hydroxide is 1:1. This means that for every mole of acetic acid used, one mole of sodium hydroxide is required for complete neutralization.

First, calculate the number of moles of sodium hydroxide used:

Moles of NaOH = Molarity × Volume (in liters)
Moles of NaOH = 1.002 M × (11.2 mL ÷ 1000 mL/1 L)
Moles of NaOH = 0.0112 moles

Since the stoichiometry of the reaction indicates a 1:1 ratio between acetic acid and sodium hydroxide, the number of moles of acetic acid used is also 0.0112 moles.

Now, calculate the concentration of the acetic acid solution:

Concentration (Molarity) = Moles ÷ Volume (in liters)
Concentration (Molarity) = 0.0112 moles ÷ (5.1 mL ÷ 1000 mL/1 L)
Concentration (Molarity) = 2.196 M

Therefore, the concentration of the acetic acid solution is approximately 2.196 M.