Question

1). Determine the molarity of a NaOH solution when each of the following amounts of acid neutralizes 25.0 mL of the NaOH solution.

(a) 20.0 mL of 0.250 M HNO3
(b) 5.00 mL of 0.500 M H2SO4
(c) 23.2 mL of 1.00 M HCl
(d) 5.00 mL of 0.100 M H3PO4

Show calculations.

Answer 1

(a) NaOH(aq) + HNO3(aq) ----> NaNO3(aq) + H2O(aq) mole of HNO3= 20 ml * 0.250/1000 = 0.005 mole 1 mole of NaOH = 1 mole of HNO3 0.005 = Molarity x 0.025 Molarity = 0.005 / 0.025 = 0.2 mol/ l or 0.2 M of NaOH

(b) 5.00 mL of 0.500 M H₂SO₄

2NaOH(aq) + H₂SO₄(aq) ----> Na2H2O4(aq) +2 H2O(aq)

mole of H₂SO₄= 5 ml * 0.500/1000 = 0.0025 mole

2 mole of NaOH = 1 mole of H2SO4

0.0025 = 2*Molarity x 0.025
Molarity = 0.0025 / 0.025*2 = 0.05 mol/ l or 0.05 M of NaOH

(c) 23.2 mL of 1.00 M HCl

NaOH(aq) + HCl(aq) ----> NaCl(aq) + H2O(aq) mole of HCl= 23.2 ml * 1.0/1000 = 0.0232 mole 1 mole of NaOH = 1 mole of HCl 0.0232= Molarity x 0.025 Molarity = 0.0232 / 0.025 = 0.928 mol/ l or 0.928 M of NaOH

(d) 5.00 mL of 0.100 M H₃PO₄

3NaOH(aq) + H₃PO₄ (aq) ----> Na₃PO₄ (aq) +3 H2O(aq)

mole of H₃PO₄= 5 ml * 0.100/1000 = 0.0005 mole

3 mole of NaOH = 1 mole of H₃PO₄

0.0005 = 3*Molarity x 0.025
Molarity = 0.0005 / 0.025*3 = 6.67*10^-3 mol/ l or 6.67*10^-3 M of NaOH