Question

A florist prepares a solution of nitrogen-phosphorus fertilizer by dissolving 5.66 g of NH4NO3 and 4.42 g of (NH4)3PO4 in enough water to make 22.0 L solution. What is the molarity of NH4+ in the solution?

Answer 1

Answer:

0.00725 M

Explanation:

Considering for $NH_4NO_3$

Mass = 5.66 g

Molar mass of $NH_4NO_3$ = 80.043 g/mol

Moles = Mass taken / Molar mass

So,  

Moles = 5.66 / 80.043 moles = 0.0707 moles

$NH_4NO_3$ will dissociate as:

$NH_4NO_3\rightarrow NH_4^++NO_3^-$

Thus 1 mole of $NH_4NO_3$ yields 1 mole of ammonium ions. So,

Ammonium ions furnished by $NH_4NO_3$ = 1 × 0.0707 moles = 0.0707 moles

Considering for $(NH_4)_3PO_4$

Mass = 4.42 g

Molar mass of $(NH_4)_3PO_4$ = 149.09 g/mol

Moles = Mass taken / Molar mass

So,  

Moles = 4.42 / 149.09 moles = 0.0296 moles

$(NH_4)_3PO_4$ will dissociate as:

$(NH_4)_3PO_4\rightarrow 3NH_4^++PO_4^{3-}$

Thus 1 mole of $NH_4NO_3$ yields 3 moles of ammonium ions. So,

Ammonium ions furnished by $(NH_4)_3PO_4$ = 3 × 0.0296 moles = 0.0888 moles

Total moles of the ammonium ions = 0.0707 + 0.0888 moles = 0.1595 moles

Given that:

Volume = 22.0 L

So, Molarity of the $NH_4^+$ is:

Molarity = Moles / Volume = 0.1595 / 22 M = 0.00725 M