Question

1.What mass of NaOH(s) must be added to 300 mL of HCl 0.25 M in order to completely neutralize this acid?
2.During a lab, you mix 2 solutions: a 100 ml solution containing 0.40 g of NaOH and a 100 mL solution containing 0.73 g of HCl. What is the concentration of H+ ions in the new solution?
3.Calculate the [H+] of a solution obtained by mixing 1 L of hydrochloric acid 1.0 M with 1 litre of sodium hydroxide 0.990 M.

Answer 1

1. mols HCl = M x L = ?
mols NaOH = mols HCl
mols NaOH = grams/molar mass. You know molar mass and mols, solve for grams.

2.mols NaOH = grams/molar mass = estimated 0.01
mols HCl = grams/molar mass = estimated 0.02.
NaOH + HCl ==> NaCl + H2O
0.02 mols HCl - 0.01 mols NaOH = 0.01 mols HCl unreacted.
M HCl = (H^+) = mols HCl/L solution. (L soln is 0.100 + 0.100 = ?)

3. mols HCl = M x L = 1.00
mols NaOH = M x L = 0.990
M HCl = (H^+) = (1.00-0.990)/2.00L = ?

Answer 2

Explain how you would prepare 0.25m of nitric acid in a 500ml volumetric flask