gammagirl

yes should if be (x +.00622)(4x^2)? Idk how the solution to that math.

Determine the mass in grams of lead (II) iodide that will dissolve in 500.0 mL of a solution containing 1.03 grams of lead (II) nitrate. Ksp of lead (II) iodide is 1.4 x 10-8. 1.03 g x mol = .00622 mol ------ ------ ------------- 0.5 L 331.2 L PbI2 (s) ==> PbI2(aq) ==>Pb2+ + 2I- .00622 2x^2 (.00622)(2x)^2=1.4 x 10^-8 x = 7.5 x 10^-4 M 7.5 x 10^-4 m/L x .5 L x 461.01 g/mole = .173 g

Absorbance of FeScn2+ And the link IS the experiment And that is the point the reaction is in equilibrium but the questions are confusing. If The Fe3+ was in excess, then Scn-=FeScn2+. But for the first question question, Fe 3+=Scn - is the same amount (as in the lab). It is something intuitive and easy.

1.Consider a reaction mixture that has initial concentrations of Fe3+ = 0.0050 M and SCN– = 0.0050 M. Without doing any calculations, which of the following values do you know? a) The equilibrium concentrations of Fe3+, SCN– , and FeSCN2+ b) The sum of the equilibrium concentrations of Fe3+ , SCN– , and FeSCN2+ c) The product of the equilibrium concentrations of Fe3+ , SCN– , and FeSCN2+ d) The ratio of equilibrium concentrations of products to reactants: [FeSCN2+]/[Fe3+][SCN– ] e) The ratio of equilibrium concentrations of reactants to products: [Fe3+][SCN– ]/[FeSCN2+] Explain your answer, Earlier in the experiment, a calibration curve measuring absorbance on the y-axis and concentration on the x-axis was generated from a set of 3 standard solutions. So, I am thinking perhaps the d and e are known due to a concentration/absorbance ratio that graph? Next is this question, Consider a reaction mixture that has an initial concentration of FeSCN2+ = 0.0050 M, no Fe3+ or SCN. Without doing any measurements or calculations, which of the following two values do you know? a) The equilibrium concentrations of Fe3+, SCN– , and FeSCN2+ b) The sum of the equilibrium concentrations of Fe3+, SCN– , and FeSCN2+ c) The product of the equilibrium concentrations of Fe3+, SCN– , and FeSCN2+ d) The ratio of equilibrium concentrations of products to reactants: [FeSCN2+]/[Fe3+][SCN– ] e) The ratio of equilibrium concentrations of reactants to products: [Fe3+][SCN– ]/[FeSCN2+] Explain your answer. By the way the equation for both is , the equation is Fe3+ (aq) + HSCN (aq) <-----> FeSCN2+ (aq) and everything is 1:1. Using the ice table, Fe3+ is 0.0050M -x , SCN-, is 0.0050M- x. So, we know a and b?

What do you mean by slightly different noise?

Is the bottom line that both values are nearly perfect straight lines making the differences between the numbers negligible?

How is a correlation coefficient of 0.999992 better than a correlation coefficient of 0.99996 even though both analyses resulted in the same molar absorptivity for the analyte using spectrophotometric analysis? Does 0.999992 result in a higher correlation coefficient than 0.99996? If so, why?

I was thinking "A shoulder band usually appears on the lower wavenumber side in primary and secondary liquid amines arising from the overtone of the N–H bending band: this can confuse interpretation."

When obtaining an IR spectrum, a secondary amine should exhibit only one N-H stretch, however, when a student ran his sample, two N-H stretches appeared in the spectrum. What could account for this?

Both are good points. Thank-you

A secondary amine should only exhibit only one N-H stretch. However, 2 N-H stretches appeared in the spectrum. What could account for this?

I used 3 mL of 95% ethanol? 3mL x 0.789/ml x mol/46.068 g = 0.0513gm OR 95gm/100ml =x/3ml, x= 2.85 gm ethanol x mol/46.068 g= 0.06187 gm

Thank-you

The answers mentioned above are correct, but I thoroughly understand the experiment at a higher level. Therefore, the Henderson-Hasselbach case is closed.

I appreciate you.

Determine the solution to the following differential equation: dA/dt = -0.3A and A(2) = 400

Part 1: (1) Made an unbuffered stock solution of DPH-diphenhydramine (100mg/100ml) in water (2) Made ph4 buffered stock solution of DPH (3) Made ph7 buffered stock solution of DPH (4) Made a ph10 buffered stock solution of DPH Part 2: Obtain absorbance at 252 of solutions in Part 1 Part 3: Add 5ml of DPH stock solutions to 5 ml hexanes and obtain absorbances. question: even though a buffered solution at pH7 should have a 100:1 ratio of ammonium: amine, why does the absorbance of the aqueous solution after extraction have such a large difference when compared to the unextracted stock solution of DPH of ph7? question: Assuming molar absorptivity of DPH is 388L/mole cm (a) calculate the approximate conc of DPH in each stock solution, before and after extraction with hexane. (did it) (b) Why don't these values match the expected ratios based on Henderson-Hasselbach equation? (my answer: HH is valid when it contains equilibrium concentrations of an acid and conjugate base. In this lab, changing pH increases the amount of DPH moving to the organic layer from the aqueous layer, changing the expected ratios of conjugated base and acid.)

I have another question. Even though a buffered solution at pH 7 should have a 100:1 ratio of ammonium ion: amine, why does the absorbance of the aqueous solution after extraction with hexane have such a large difference when compared to the unextracted stock solution of the amine a ph 7?

Step 1: Creating pH 4, 7, 10 buffered stock solutions. Step 2: Obtain absorbance by spectroscopy of buffered solutions. Step 3: Obtain absorbance of stock solutions after extraction with hexanes. Step 4: calculate concentration with Beer's Law Question: Why don't these values match the expected ratios based on the Henderson-Hasselbalch equation? The difference is due to the buffer solutions are able to withstand most changes in pH and maintain a reasonably stable pH?

When extracting diphenhydramine from hexane and calculating the concentration of diphenhydramine using the Beer-Lambert law, the ratios of NH3/NH4+ don't match the expected ratios based on the Henderson-Hasselbalch equation. Why?

Only someone really smart could give an answer like that, especially since no solubility data was given in cold versus hot to hint at that issue as the main point. That explanation is the entire basis of recrystallization.

The property you are exploiting is removal of the impurities. And why is always the same......is the puzzle to me. There are impurities that are soluble and insoluble in the mother filtrate along with the pure product.

Dear hypervalent_iodine, if you start off with a 100gm mass 10gm is impurity 90 gm is pure product. So during the second crystallization, at most 80 gm is the maximum yield. Then, ......

Gentlemen, The vague answer, possibly, is that some crystals are left behind in the solvent during each recrystallization so this causes a decrease in recovery. In addition, successive recrystallizations result in soluble impurities contaminating the filtrate, which reduces the yield of pure crystals. The question is," Is there some math that goes along with this 10 % impurity that results in successive 20% decrease in yield with each subsequent crystallization?"