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 $(NH_4)_3PO_4$.

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

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 $3\times 0.03=0.09$ 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.

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

$\text{Concentration of }NH_4^+=\frac{0.09mole}{0.06L}=1.5M=0.15E^1M$

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