Question

urgent!!!

The neutralization reaction between ammonium hydroxide (NH4OH, MM = 35.04 g/mol) and phosphoric acid (H3PO4, MM = 98.00 g/mol) proceeds according to the following balanced eqation: 3NH4OH(aq) + H3PO4(aq) ->3H20(1) + (NH4)3PO4(aq) What is the final molarity of the ammonium ion (NH4+, MM = 18.05 g/mol) when 25.0 mL of 1.20 molar ammonium hydroxide is reacted with 25.0 mL of 2.00 molar phosphoric acid?

Answer 1

no of moles of NH4OH   = molarity * volume in L

                                   = 1.20*0.025   = 0.03 moles

no of moles of H3Po4 = molarity * volume in L

                                 = 2*0.025   = 0.05moles

3NH4OH(aq) + H3PO4(aq) -------------> (NH4)3PO4(aq) + 3H2O (l)

1 moles of H3Po4 react with 3 moles of NH4OH

0.05 moles of H3PO4 react with = 3*0.05/1    = 0.15 moles of NH4OH is required

NH4OH is limiting reactant

3 moles of NH4OH react with excess of H3PO4 to gives 1 mole of (NH4)3PO4

0.03 moles of NH4OH react with excess of H3Po4 to gives 0.03*1/3 = 0.01 moles of (NH4)3Po4

total volume of solution = 25 + 25 = 50ml   = 0.05L

molarity of (NH4)3Po4 = no of moles/total volume in L

                                = 0.01/0.05   = 0.2M

(NH4)3PO4(aq) -------------> 3NH4^+ (aq) + PO4^3- (aq)

0.2M --------------------------- 3*0.2M

molarity of NH4^+   = 0.6M >>>>>answer