6.0 of ammonium phosphate is dissolved into water to produce

M[(NH4)3PO4] = 149 g/mol
m[(NH4)3PO4] = 6 g
n[(NH4)3PO4] = m[(NH4)3PO4]/M[(NH4)3PO4] = 6 g/149 g/mol = 0.04 mol
V[(NH4)3PO4(ag)]=300 mL=0.3 L
(NH4)3PO4 = 3NH4^+ + PO4^3-
c(NH4^+) = n(NH4^+)/V[(NH4)3PO4(ag)] = 3*0.04 mol / 0.3 L = 0.4 mol/L
c(PO4^3-) = n(PO4^3-)/V[(NH4)3PO4(ag)] = 0.04 mol / 0.3 L = 0.13 mol/L

写出离子式，然后用C1V1=C2V2求就行了。

首先楼主，不是（NH4）2PO4，而是（NH4）3PO4，首先列平衡方程
(NH4)3PO4→3NH4+P04
NH4离子concentration：6g× 1mol/149.12g ×
3mol of NH4/1mol of (NH4)3PO4× 1mol/0.3L=0.40MOL/L
PO4离子concentration：6g× 1mol/149.12g × 1mol of PO4/1mol of (NH4)3PO4× 1mol/0.3L=0.13MOL/L