In an acid base titration 32.5 mL of sodium hydroxide, NaOH, were neutralized by 17.6 mL of 0.180 mol/L sulfuric acid, H2SO4. Calculate the concentration of the sodium hydroxide.
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
mols H2SO4 = M x L = ?
mols NaOH = twice that of H2SO4.[Note the coefficients in the balanced equation.]
M NaOH = mols NaOH/L NaOH
To calculate the concentration of the sodium hydroxide, we can use the concept of stoichiometry.
The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4) is:
2 NaOH + H2SO4 → Na2SO4 + 2 H2O
From the balanced equation, we can see that one mole of sulfuric acid reacts with two moles of sodium hydroxide.
First, we need to determine the number of moles of sulfuric acid used in the reaction. We can do this by using the equation:
moles of sulfuric acid = concentration × volume
Given that the volume of sulfuric acid used is 17.6 mL and the concentration of the acid is 0.180 mol/L:
moles of sulfuric acid = (0.180 mol/L) × (17.6 mL / 1000 mL/L) = 0.003168 mol
Since the reaction is 1:2 ratio of sodium hydroxide to sulfuric acid, we can determine the number of moles of sodium hydroxide:
moles of sodium hydroxide = 0.003168 mol × (1 mol NaOH / 2 mol H2SO4) = 0.001584 mol
Now, we can calculate the concentration of the sodium hydroxide by using the equation:
concentration = moles / volume
Given that the volume of sodium hydroxide used is 32.5 mL:
concentration = 0.001584 mol / (32.5 mL / 1000 mL/L) ≈ 0.0487 mol/L
Therefore, the concentration of the sodium hydroxide is approximately 0.0487 mol/L.