13.In a study of rust-removing solutions, 27.8 mL of 0.115 mol/L phosphoric acid reacts completely with 0.245 mol/L sodium hydroxide. Predict the minimum volume of sodium hydroxide required for this reaction?
3 NaOH + H3Po4 = 3H20 + Na3PO4
(27.8 ml)(0.115) = 0.003197 mol H3Po4
0.003197 * (3 mol NaOh) = 0.009591 mol
v = 0.009591 mol/ 0.245 mol/L
v = 39.1 ml
You made two error in naming but they cancel.
27.8 x 0.115 is 0.003197 MILLImoles (not mols) BUT
at the end you divide millimoles/mL and that is M. Your answer of 39.1 mL is right.
To predict the minimum volume of sodium hydroxide required for this reaction, we need to use the balanced chemical equation and the molar ratios between the reactants.
From the balanced chemical equation:
3 NaOH + H3PO4 → 3 H2O + Na3PO4
We can see that the molar ratio between NaOH and H3PO4 is 3:1. This means that for every 3 moles of NaOH, 1 mole of H3PO4 is reacted completely.
First, let's calculate the number of moles of H3PO4 in the given volume:
Moles of H3PO4 = volume (in liters) x concentration (in mol/L)
Moles of H3PO4 = (27.8 mL / 1000 mL/L) x 0.115 mol/L
Moles of H3PO4 = 0.003197 mol
Next, using the molar ratio, we can determine the number of moles of NaOH required for the reaction. We want to find the minimum volume, so we calculate based on the given concentration of NaOH (0.245 mol/L):
Moles of NaOH = 0.003197 mol H3PO4 x (3 mol NaOH / 1 mol H3PO4)
Moles of NaOH = 0.009591 mol
Finally, we can calculate the minimum volume of sodium hydroxide required using its concentration:
Volume of NaOH = moles of NaOH / concentration of NaOH
Volume of NaOH = 0.009591 mol / 0.245 mol/L
Volume of NaOH ≈ 39.1 mL
Therefore, the minimum volume of sodium hydroxide required for this reaction is approximately 39.1 mL.