Question

Not yet answered Marked out of 2.00 What is the concentration of NH4+ in 60.0 mL of a 0.50 M solution of (NH4)3PO4? (To write your answer using scientific notation use 1.0E-1 instead of 1.0 x 10-?) Answer: Answer

Answer 1

Answer : The concentration of
`NH_4^+` ion is
`0.15E^1M`

Explanation :

First we have to calculate the moles of
`(NH_4)_3PO_4`.

`\text{Moles of }(NH_4)_3PO_4=\text{Concentration of }(NH_4)_3PO_4* \text{Volume of solution}=0.50M* 0.06L=0.03mole`

The balanced chemical reaction will be:

`(NH_4)_3PO_4\rightleftharpoons 3NH_4^++PO_4^(3-)`

From the reaction we conclude that,

1 mole of
`(NH_4)_3PO_4` dissociate to give 3 moles of
`NH_4^+` ion and 1 mole of
`PO_4^(3-)` ion

So,

0.03 mole of
`(NH_4)_3PO_4` dissociate to give
`3* 0.03=0.09` moles of
`NH_4^+` ion and 0.03 mole of
`PO_4^(3-)` ion

Now we have to calculate the concentration of
`NH_4^+` ion.

`\text{Concentration of }NH_4^+=\frac{\text{Moles of }NH_4^+}{\text{Total volume}}`

`\text{Concentration of }NH_4^+=(0.09mole)/(0.06L)=1.5M=0.15E^1M`

Therefore, the concentration of
`NH_4^+` ion is
`0.15E^1M`