Question

2.poets.] Phosphoric acid (75.0 m1L of 0.250 M) is added n hydroxide solution. mt oro.250 M) is added to 425.00 mL or o.osso M barium a. Write a balanced chemical b. Identify the spectator ions. equation for the above reaction rnte a net-ionic equation for the above reaction. Calulate the concentration of the excess H or OH ions left in this solution.

Answer 1

a) 2H3PO4_(aq) + 3Ba(OH)2_(aq) ----------> Ba3(PO4)2_(s) + 6H2O_(l) balanced reaction.

b) 6H⁺_(aq) + 2PO4^-3_(aq) + 3Ba⁺²_(aq) + 6OH^-_(aq) ---------->  Ba3(PO4)2_(s) + 6H2O_(l)   

no ions present bothsides so no spectator ions

c) 6H⁺_(aq) + 2PO4^-3_(aq) + 3Ba⁺²_(aq) + 6OH^-_(aq) ---------->  Ba3(PO4)2_(s) + 6H2O_(l) net-ionic equation

d) moles of H3PO4 = 0.250 x 75.0/1000 = 0.01875

moles of Ba(OH)2 = 0.055 x 425 /1000 = 0.023375

according to balanced reaction

3 moles Ba(OH)2 reacts with 2 moles H3PO4

0.023375 moles Ba(OH)2 reacts with 0.023375 x 2 / 3 = 0.01558 moles H3PO4

exess H3PO4 left = 0.01875 - 0.01558 = 0.00317 moles

total volume = 425 + 75 = 500 mL = 0.50 L

[H3PO4] left = 0.00317 / 0.5 = 0.0634 M

exess H⁺ left = 0.0634 x 3 = 0.1902 M

exess H⁺ left = 0.1902 M