[Buddy_123]

Question: A student mixes 5.00mL of 2.00x10^-3M Fe(NO3)3 with 5.00mLof 2.00x10^-3M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN²⁺ is 1.40x10^-4M. Find K for reaction Fe³⁺(aq)+SCN^-(aq)=FeSCN²(aq).

MY work: Fe³⁺=(.005)(2.00x10^-3)=1x10^-5

1x10^-5-1.40x10^-6=8.6x10^-6
SCN^-=1x10^-5 & 8.6x10^-6

*Both Fe and SCN are the same numbers.
M=8.6x10^-6/.01=0.00086

K==1.4x10^-4/(0.00086)²
K=189

Can someone let me know if it looks right? I'm not sure if the number is too big and I want to make sure it is correct before I work on part B. Thank You!

[AWK]

Good result.
Calculations can be shortened - dilution twice, concentration two times lower.

[Orcio_87]

Question: A student mixes 5.00mL of 2.00x10-3M Fe(NO3)3 with 5.00mLof 2.00x10-3M KSCN. She finds that in the equilibrium mixture the concentration of FeSCN2+ is 1.40x10-4M. Find K for reaction Fe3+(aq)+SCN-(aq)=FeSCN2(aq).

Initial concentration of Fe³⁺ and SCN^- (after adding solutions altogether) = 1 x 10^-3 M

Concentration of Fe³⁺ and SCN^- after reaction:

1 x 10^-3 M - 0,14 x 10^-3 M = 0,86 x 10^-3 M

Thus

K = 0,14 x 10^-3 M / [0,86 x 10^-3 M]² = 189

Good answer.