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

0.00725 M

Step-by-step 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