Question

I already know the answer to part B, i just need help figuring out how to do part C, please show me the steps.(:

Part B. What is the pH of the solution prepared by allowing 3.65 g of Na2O to react with 450.0 mL of water? Assume that there is no volume change answer :13.418

Part C,How many milliliters of 0.0100 M HCl are needed to neutralize the NaOH solution prepared in the part B?

Answer 1

Answer:

Na₂O reacts with H₂O to forms NaOH

The reaction will be Na₂O + H₂O → 2NaOH

Hence, 1 mole of Na₂O will results in the formation of 2 moles of NaOH (2 OH^- ions)

Moles of Na₂O = 3.65/61.979 = 0.05889 moles

Moles of OH^- will be produced = 0.11778 moles and molarity = 0.11778/0.450 = 0.2617 M

From relative relation;

M_NaOH.V_NaOH = M_HCl.V_HCl => V_HCl = (0.2617 M x 450 mL)/0.01 M = 11776.5 mL of 0.01 M HCl solution will be needed for the neutralization of NaOH prepared in part B.

Please let me know, if you have any doubt by commenting below the answer.

Thanks