Question

What is the final NH4+ concentration if 23.0 mL of a 2.59 M NH4F solution is added to 88.2 mL of a 1.73 M (NH4)3PO4 solution?

Answer 1

first we must consider this reaction
(NH4)3PO4(aq) ---> 3 NH4+(aq) + PO4---(aq)
NH4F-> NH4+ + F-

hence 1 mole of (NH4)3PO4 and 1 mole of NH4F makes 3 and 1 mole of NH4+ respectively

so ,moles of (NH4)3PO4= 1.73 * 0.0882 = 0.153

so ,moles of NH4F= 2.59*0.023 = 0.0596

hence ,moles of NH4+ ions = 0.153*3 + 0.0596 = 0.5186= 0.52 moles

concentration of NH4+ = 0.52/ (23+88.2) *(10^-3) = 4.676M