Question

Acid Base dissociation

		molar mass	mass	volume of H2O	Conc of Solution (M)	Theoritical pH
NaH2PO4		120g/mol	.6g	30ml	?	?
Na2HPO4*7H2O		268g/mol	.6g	30ml	?	?
Na3PO4*12H2O		380g/mol	.2g	10ml	?	12

Answer 1

for NaH₂PO₄ 1 molarity = 120 g / 1L

therefore for 0.6 g = 0.6/120 = 0.005 M

for 1000 ml of water in 0.6 g of NaH₂PO₄ = 0.005 M

for 30 ml = 0.005 x 10 / 1000 = 0.00005 M

pH = -log [H⁺] = log [0.00005] = 3.82

for Na2HPO4*7H2O 1 molarity = 268 g / 1L

therefore for 0.6 g = 0.6/268 = 0.0022 M

for 1000 ml of water in 0.6 g of Na2HPO4*7H2O = 0.0022 M

for 30 ml = 0.0022 x 30 / 1000 = 0.0000671 M

pH = -log [H⁺] = log [0.0000671] = 4.17

for Na3PO4*12H2O 1 molarity = 380 g / 1L

therefore for 0.2 g = 0.2/380 = 0.000526 M

for 1000 ml of water in 0.2 g of Na3PO_4.12H2O = 0.000526 M

for 30 ml = 0.000526 x 10 / 1000 = 0.00000526 M

pH = -log [H⁺] = log [0.00000526] = 5.27