Question

To obtain [FeSCN2+]eq in tube 5, you made the assumption that 100% of the ions SCN– had reacted. Now that you know the value for Kc (average in table 5), build an ICE diagram to calculate the true percentage of SCN– reacted in tube 5

I have my Kc value but don't know how to implement this to find my true percentage.
Thank you

Answer 1

Fe³⁺ +SCN-↔Fe(SCN)]²⁺

Kc=1.12*10^-3

If concentration [Fe3+]=x

[SCN-]=y

	[Fe3+]	[SCN-]	[Fe(SCN)]2+
Initial	x	y	0
change	x-a	y-a	a
equilibrium	x-a	y-a	a

Kc=[Fe(SCN)]²⁺/[Fe3+] [SCN-]=a /(x-a) (y-a)=a/xy [neglecting a <<<x,y]

1.12*10^-3= a/xy…………….(1)

% of SCN- reacted=a/y *100 ...........(2)

Plug in the initial concentration of [Fe3+]=x and [SCN-]=y in equation (1) to calculate ‘a’ =equilibrium concentration

Then put the value in equation (2) to calculate true % of SCN- reacted.