If 33.0 mL of 0.002 M aqueous H3PO4 is required to neutralize 28.0 mL of an aqueous solution of NaOH, determine the molarity of the NaOH solution.
H3PO4 + 3NaOH ==> 3H2O + Na3PO4
mols H3PO4 = M x L = ?
mols NaOH = mols H3PO4 x (3 mols NaOH/1 mol H3PO4) = moles H3PO4 x 3 = ?
M NaOH = mols NaOH/ L NaOH
To determine the molarity of the NaOH solution, we can use the concept of stoichiometry and the equation for the neutralization reaction between H3PO4 and NaOH:
H3PO4 + 3NaOH -> Na3PO4 + 3H2O
First, let's find the moles of H3PO4 in the given volume (33.0 mL) and molarity (0.002 M). We can use the formula:
moles = volume (L) x molarity
First, we need to convert the volume from milliliters to liters:
33.0 mL = 33.0 mL / 1000 mL/L = 0.033 L
Now, calculate the moles of H3PO4:
moles of H3PO4 = 0.033 L x 0.002 M = 0.000066 mol
According to the balanced equation, 1 mole of H3PO4 reacts with 3 moles of NaOH. Therefore, the number of moles of NaOH in the reaction is also 0.000066 mol.
Next, let's calculate the molarity of the NaOH solution. We'll use the formula:
molarity = moles / volume (L)
The volume of the NaOH solution is given as 28.0 mL, so we convert it to liters:
28.0 mL = 28.0 mL / 1000 mL/L = 0.028 L
Now, calculate the molarity of the NaOH solution:
molarity of NaOH = 0.000066 mol / 0.028 L ≈ 0.00236 M
Therefore, the molarity of the NaOH solution is approximately 0.00236 M.