Question

27.5 mL of a 0.250 M ammonium carbonate solution is added to 36.0 mL of 0.1250 M calcium chloride. What is the concentration of ammonium and calcium ions in the final solution?

1. ammonium ion:_____M
2. calcium ion:_____M

Answer 1

Here is your answer.

Hope you will understand.

For ammonium carbonate,

Taken volume of ammonium carbonate(V1) = 27.5mL

And concentration (S1)= 0.250M

Total volume after calcium chloride added(V2)= (27.5+36.0) mL = 63.5 mL

We know, V1×S1 = V2×S2

So, putting the values of V1, S1,V2 , we will get the concentration of ammonium carbonate after addition of calcium chloride.

So, 27.5 mL × 0.250M = 63.5 mL × S2

or, S2 = 0.1082677 M

So, concentration of ammonium carbonate is 0.1082677 M.

We can see from 1 mole ammonium carbonate, we get 2 mole ammonium ion, because

(NH4)2CO3 --------> 2NH4+ + CO32-

So, the concentration of ammonium ion = 2 × concentration of ammonium carbonate

= 2 × 0.1082677 M = 0.2165354 M (Answer)

For calcium chloride,

Taken volume of calcium chloride (V1)= 36.0 mL and concentration (S1) = 0.1250 M

Total volume (V2) =(27.5 +36.0) mL =63.5 mL

We know, V1× S1 = V2 × S2

or, 36.0 mL× 0.125 M = 63.5 mL × S2

or, S2 = 0.070866 M

We get 1 mole calcium ion from 1 mole calcium chloride,

CaCl2 -----------> Ca2+ + 2Cl-

So, concentration of calcium ion is 0.070866 M.(answer)

Thank you for reading. ?