In an experiment 5 00 mL of a 3.80 times 10^-3 M aqueous solution of Co(NO_ 3)_2

Question

In an experiment 5 00 mL of a 3.80 times 10^-3 M aqueous solution of Co(NO_ 3)_2 is initially combined with 10 00 mL of a 900 times 10^-3 M aqueous solution of KSCN. The solution was then diluted lo a total volume of 25 00 mL by addition of solvent. It was determined that the absorbance. A, in the equilibrium solution was equal to 0 302. Complete the table below with NUMERICAL VALUES, (not functions of x.) in scientific notation. (If you can't find the equilibrium molarity of [Co(SCN)_4]^2- assume a FAKE value of 3.24 times 10^-4 m)

Explanation / Answer

Ok to find the moles of each compound you need to organize your data and use it in the equation of Molarity.

M = n /V(L)

n = MxV(L)

so you have initially Co(NO3)2   5.0 mL = 5x10-3L and 3.80x10-3 M (the numbers are not really clear so i will use the quantity i think it is, if there is any change please do the math with the real value)

and KSCN 10 mL = 0.01 L and 9x10-2 M (i can¨t see clearly the little number)

we calculate the moles of each compund.

n Co(NO3)2 = 5x10-3 L (3.80x10-3M) = 1.9x10-5 mol.

n KSCN = 0.01 L (9x10-2 M) = 9x10-4 mol

As you can see the moles depend on the volume of the solution, then if we do a dilution it will change the moles.

Our new volume is 25 mL = 0.025 L the we recalculate the moles of each compound.

n Co(NO3)2 = 0.025 L (3.80x10-3M) = 9.5x10-5 mol.

n KSCN = 0.025 L (9x10-2 M) = 2.25x10-3 mol

Now we have that the absorbance of A = 0.302 and e of Co(SCN)4=437 L/mol*cm

A = e*b*c, where 'e' is the molar absorptivity, 'b' is how much distance the light source passes through the solution in centermeters, 'c' is the concentration of the solution in mol/L

for b we will asume 1 cm.

then we can find c = A / e * b

c(Co(SCN)4) = 0.302 /(1cm) (437 L/mol*cm)

c(Co(SCN)4) = 6.9x10-4 mol/ L = molarity in equilibrium.

Then to calculate the moles of Co(SCN)4 we have the molarity and the volume = 25 mL = 0.025 L

n = 6.9x10-4 mol/L * (0.025L)

n Co(SCN)4 = 1.7x10-5 mol.

Now we now that in equilibrium the moles of Co(SCN)4 formed we have to substract it to the initial moles of the compounds.

n Co(NO3)2 = 9.5x10-5 mol. - 1.7x10-5 mol. = 7.8x10-5 mol

n KSCN = 2.25x10-3 mol - 1.7x10-5 mol.= 2.23x10-3 mol

now we can find the molarity of each of the compunds in equilibrium with the 25 mL volume

MCo(NO3)2 =  7.8x10-5 mol / 0.025 L = 3.12x10-3 M

MKSCN = 2.23x10-3 mol / 0.025 L = 0.089 M

To find the K

K = [Co(SCN)4] / [KSCN ]*[Co(NO3)2 ]

K = (6.9x10-4 ) / (0.089 )*(3.12x10-3 )

K = (6.9x10-4 ) / (2.7x10-4 )

K = 2.48

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