Question

3. Use the provided experimental data to calculate the equilibrium constant, K, for the reaction given below. Note the stoichiometry is not all 1 to 1. (7) Include the units for K Ke=? Co3+(aq) + 2SCN-(aq) = Co(SCN)2(aq) 5.0 ml of 0.10 M CO(NO3)3 was mixed with 5.0 ml of 0.050 M KSCN The absorbance of the equilibrium mixture at 491 nm was 0.157. The linear regression equation for the (Co(SCN)2-) calibration curve was: y = (4.3 x 103)x + 0.0074 units for K:

Answer 1

First we can obtain the final concentration of Co(SCN)₂⁺using the linear regression equation:

$y=(4.3*10^{3})x+0.0074$

that means:

$absorbance=(4.3*10^{3})*concentration+0.0074$

so:

$[Co(SCN)_2^{+}]=concentration=\frac{absorbance-0.0074}{4.3*10^{3}}$

$[Co(SCN)_2^{+}]=\frac{0.157-0.0074}{4.3*10^{3}}=3.48*10^{-5}M$

We need to know the number moles of product (that will help us to know the number of moles of each reagent that reacted and then, their final concentration)  using unit factor. As we have the concentration, and the final volume is the sum of the volumes of reagents (5mL+5mL=10mL)

$n_{[Co(SCN)_2^{+}]}=(10mL)(\frac{3.48*10^{-5}M}{1,000mL})=3.48*10^{-7}mol$

With this value and the stoichiometric numbers we can know the number of moles of each reagent that reacted:

$n_{Co(NO_3),reac}=(3.48*10^{-7}mol_{[Co(SCN)_2^{+}]})(\frac{1mol_{Co^{3+}}}{1mol_{[Co(SCN)_2^{+}]}})=3.48*10^{-7}mol_{Co^{3+}}$

$n_{SCN^{-},reac}=(3.48*10^{-7}mol_{[Co(SCN)_2^{+}]})(\frac{2mol_{SCN^{-}}}{1mol_{[Co(SCN)_2^{+}]}})=6.96*10^{-7}mol_{Co^{3+}}$

Now we have to obtain the initial number of moles of each reagent:

$n_{Co(NO_3)}=(5mL)(\frac{0.10mol}{1,000mL})=5*10^{-4}mol$

$n_{KSCN}=(5mL)(\frac{0.050mol}{1,000mL})=2.5*10^{-4}mol$

We are ready to calculate the final number of moles of each reagent (at equilibrium):

$n_{Co(NO_3)}=5*10^{-4}mol-3.48*10^{-7}mol=4.965*10^{-4}mol$

$n_{KSCN}=2.5*10^{-4}mol-6.96*10^{-7}mol=2.493*10^{-4}mol$

As the final volume is 10 mL, we can obtain the final concentrations:

$M_{Co(NO_3)}=(\frac{1,000mL}{1L})(\frac{4.965*10^{-4}mol}{10mL})=0.0496M$

$M_{KSCN}=(\frac{1,000mL}{1L})(\frac{2.4930*10^{-4}mol}{10mL})=0.025M$

Now we can substitute in the equilibrum expression:

$Kc=\frac{[Co(SCN)_2^{+}]}{[Co^{3+}][SCN^{-}]^{2}}=\frac{3.48*10^{-5}M}{(0.0496M)(0.025M)^{2}}=1.12$