Question

Give the concentration of each ion present in the solution formed by mixing together these two solutions desribed. 51.7 mL of 0.175 M Fe(NO3)3 and 23.7 mL of 0.509 M CaCl2

Answer 1

Given :

51.7 mL of 0.175 M Fe(NO3)3 and 23.7 mL of 0.509 M CaCl2

Volume of Fe(NO3)3 = 51.7 mL = 0.0517 L

[Fe(NO3)3] = 0.175 M

Volume of CaCl2 = 23.7 mL = 0.0237 L

[CaCl2]= 0.509 M

Calculation of moles :

Mol Fe(NO3)3 = Molarity x volume in L

= 0.175 M x 0.0517

= 0.009048 mol

Mol CaCl2 = 0.509 M x 0.0237 L

= 0.0121 mol CaCl2

Reaction between Fe(NO3)3 and CaCl2

Fe(NO3)3 (aq) + CaCl2 (aq)   ---- > Ca(NO3)2 (aq) + FeCl2 (aq)

If we see the reaction all are in aqueous phase so this reaction is not possible

No any ions in the solution will react. We calculate concentration of each ions.

Fe(NO3)3

1). Fe³⁺

[Fe³⁺] = mol of Fe³⁺ / volume in L

Here we use total volume in L

Total volume = 0.0517 L + 0.0237 L = 0.0754 L

Number of moles of Fe³⁺ = Number of moles of Fe(NO3)3 = 0.009048 mol

[Fe³⁺] = 0.009048 mol / 0.0754 L = 0.1199 M = 0.120 M

[NO₃^-]

Mol of NO₃^- = moles of Fe(NO3)3 x 3 mol NO₃^- / 1 mol Fe(NO3)3

= 0.009048 x 3 mol NO₃^-

= 0.0271 mol / 0.0754 L

= 0.3599 M

= 0.360 M

[Ca²⁺]

Mol of Ca²⁺ = mol CaCl2 = 0.0121 mol Ca²⁺

[Ca²⁺ ] = 0.0121 mol / 0.0754 L

= 0.160 M

[Cl-]= 2 x mol CaCl2 / 0.0754 L

= 2 x 0.0121 mol / 0.0754 L

= 0.0241 / 0.0754 L

= 0.320 M