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Sorry, since it is unknown of what experiment or laboratory analysis you're talking about, this question is unanswerable. It also is not accurate enough: FeNCS is not a good formula, SCN is an anion: SCN- and the sentence:

".... when the calibration curve was prepared(??) would this raise or lower the value of Keq" is difficult to interprete as such a curve is not adequately described.

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Why can you use a calibration curve for getting equilibrium data?

I'm doing this lab, and it was explained to me by my instructor... Basically, on the x-axis you'll have the concentration of a substance, and on the y-axis you'll have the %T, or A, the absorbance of the substance when it's put into a spectrophotometer. so you plot the points, get a line of best fit (this is your calibration curve), and then basically you use that line to get the concentration of the substance, when you've already calculated the A. . And from that concentration, you can extrapolate the concentration of the reactants/products (Depending on what you're looking for) to find the equilibrium constant. Here's an example: iron and thiocyanate ions bond to form iron thiocyanate in the following equation: Fe(3+) + SCN(-) --> FeNCS(2+) For the experiment I did, a calibration curve was made with reacted Fe(3+) and SCN(-). So for my calibration curve, I got the concentration of FeNCS(2+) on the x-axis, and the absorbance or A on the y-axis. So you basically got to find the equilibrium concentrations of the Fe(3+) and the SCN(-), cuz you've already got the equilibrium concentration for FeNCS(2+). So you start with the initial, use the equilibrium FeNCS(2+) to calculate the equilibrium concentration of the reactants. Here's the equation: equilibrium [Fe(3+)] = initial [Fe(3+)] - equilibrium [FeNCS(2+)]. And the same goes for the SCN ion.. you just switch out the numbers. So now that you got all that, it's simply a matter of dividing the product concentrations by the multiplication of the reactant concentrations. and boom, you have found the equilibrium constant. Keep it simple stupid, y'all. There is another way to find it, and that's using the Beer-Lambert's law..


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Why can you use a calibration curve for getting equilibrium data?

I'm doing this lab, and it was explained to me by my instructor... Basically, on the x-axis you'll have the concentration of a substance, and on the y-axis you'll have the %T, or A, the absorbance of the substance when it's put into a spectrophotometer. so you plot the points, get a line of best fit (this is your calibration curve), and then basically you use that line to get the concentration of the substance, when you've already calculated the A. . And from that concentration, you can extrapolate the concentration of the reactants/products (Depending on what you're looking for) to find the equilibrium constant. Here's an example: iron and thiocyanate ions bond to form iron thiocyanate in the following equation: Fe(3+) + SCN(-) --> FeNCS(2+) For the experiment I did, a calibration curve was made with reacted Fe(3+) and SCN(-). So for my calibration curve, I got the concentration of FeNCS(2+) on the x-axis, and the absorbance or A on the y-axis. So you basically got to find the equilibrium concentrations of the Fe(3+) and the SCN(-), cuz you've already got the equilibrium concentration for FeNCS(2+). So you start with the initial, use the equilibrium FeNCS(2+) to calculate the equilibrium concentration of the reactants. Here's the equation: equilibrium [Fe(3+)] = initial [Fe(3+)] - equilibrium [FeNCS(2+)]. And the same goes for the SCN ion.. you just switch out the numbers. So now that you got all that, it's simply a matter of dividing the product concentrations by the multiplication of the reactant concentrations. and boom, you have found the equilibrium constant. Keep it simple stupid, y'all. There is another way to find it, and that's using the Beer-Lambert's law..


When 90.0mL of 0.10M Fe3 plus is added to?

When 90.0mL of 0.10M Fe3 plus 3 is added to 10.0 mL of SCN minus 1, you get an equilibrium molar concentration of FeNCS plus 2. This is determined from a calibration curve of 1.0x10-6 mol/L.