Let's solve the problem step-by-step:
1. Understand the Dissolution Reaction:
The dissolution reaction of ammonium phosphate [tex]\(\left( (NH_4)_3PO_4 \right)\)[/tex] in water is given as:
[tex]\[
(NH_4)_3PO_4(s) \rightarrow 3 NH_4^+(aq) + PO_4^{3-}(aq)
\][/tex]
This equation tells us that 1 mole of [tex]\(\left( (NH_4)_3PO_4 \right)\)[/tex] produces 3 moles of [tex]\(NH_4^+\)[/tex].
2. Determine the Given Concentration and Volume:
We are given the concentration of [tex]\(NH_4^+\)[/tex] ions, which is 0.15 M, and the volume of the solution, which is 1.0 L.
3. Calculate the Moles of [tex]\(NH_4^+\)[/tex]:
The molarity (concentration) formula is:
[tex]\[
\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}
\][/tex]
Rearranging to find the moles of [tex]\(NH_4^+\)[/tex]:
[tex]\[
\text{Moles of } NH_4^+ = \text{Molarity} \times \text{Volume}
\][/tex]
Substituting the given values:
[tex]\[
\text{Moles of } NH_4^+ = 0.15 \text{ M} \times 1.0 \text{ L} = 0.15 \text{ moles}
\][/tex]
4. Determine the Moles of [tex]\(\left( (NH_4)_3PO_4 \right)\)[/tex]:
Knowing that 1 mole of [tex]\(\left( (NH_4)_3PO_4 \right)\)[/tex] produces 3 moles of [tex]\(NH_4^+\)[/tex], we can set up the ratio:
[tex]\[
\text{Moles of } (NH_4)_3PO_4 = \frac{\text{Moles of } NH_4^+}{3}
\][/tex]
Substituting the moles of [tex]\(NH_4^+\)[/tex] calculated:
[tex]\[
\text{Moles of } (NH_4)_3PO_4 = \frac{0.15 \text{ moles}}{3} = 0.05 \text{ moles}
\][/tex]
Thus, the number of moles of [tex]\(\left( (NH_4)_3PO_4 \right)\)[/tex] required to produce a solution in which the concentration of [tex]\(NH_4^+\)[/tex] is 0.15 M for 1.0 L of solution is 0.05 moles.
Therefore, the correct answer is:
C. 0.050
