Problem reads: A standard solution of potassium dichromate is used to determine

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Problem reads: A standard solution of potassium dichromate is used to determine the molar concentration of an unkown ferrous ammonium sulfate solution by titration. The balanced titration reaction is:

A standard solution of potassium dichromate is used to determine the molar concentration of an unknown ferrous ammonium sulfate solution by titration. The balanced titration reaction is: CnO72-(aq) + 6Fe2"(aq) + 14H+(aq) 2CH.(aq) + 6Fe3"(aq) + 7H20() (a) The mass of potassium dichromate that was added to a calibrated 500-mL volumetric flask was 0.1443 g. The calibrated volume of the 500-ml volumetric flask is 499.881 mL. Calculate the molarity of the potassium dichromate solution. (b) A 25-mL sample of the unknown ferrous ammonium sulfate solution is delivered to an Erlenmeyer flask and the titration with potassium dichromate completed. The procedure was completed for three replicates. It took the indicated volume for each replicate to reach the end point of the titration. Replicate number vol. K2Cr20, mL 12.12 3 12.16 12.10 Calculate the 95% confidence interval for these three replicates. (c) Calculate the relative uncertainty of the volume of potassium dichromate required to reach the end point. Recall that in CHEM 3111 lab, all uncertainties are 95% confidence intervals. (d) The volume of the ferrous ammonium sulfate solution delivered (in step b) and titrated with potassium dichromate was 25.012. Calculate the molar concentration of the ferrous ammonium sulfate solution.

Explanation / Answer

a) Mass of dichromate = 0.1443g

Volume of the volumetric flask = 499.88 mL = 0.49988L

Molarity = moles/liter

Molecular weight of potassium dichromate = 294.185g/mol

moles = 0.1443g/294.185g/mol = 0.00049 mol

molarity = moles/liter = 0.00049g/0.4998L = 0.00098 M

b)Mean : (12.12+12.10+12.16)/3 = 12.126

Standarad deviation = 0.0249

(Sigma)mean = (sigma)/root(N) = 0.024/root (3) = 0.0138

margin of error by multiplying the standard error by 2. 0.0138 x 2 = 0.028

Computing the confidence interval by adding the margin of error to the mean from Step 1 and then subtracting the margin of error from the mean:

          12.126-0.028 = 12.098

          12.126+0.028 = 12.154

c) Relative uncertanity = (0.028/12.126)x 100 = 0.23%

d)N1V1= N2V2

1 - Ferrous Ammonium Sulfate 2-Potassium Dichromate

N1=?

V1= 25.012 mL (given value)

N2 = (0.1443g/49)/0.49988 L = 0.00589N (Equivalent weight of pot.dichromate = 49g)

V2= 12.126 mL(Average of titer values)

Substituting in the above equation, N1 = 0.00285N

The strength in g/L = N1x 392.1 (equivalent weight of ferrous ammonium sulphate) = 1.11 g/L

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