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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

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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

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