). 10 points) An iodometric titration was used to find the concentration o (MnO4
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). 10 points) An iodometric titration was used to find the concentration o (MnO4) in an iodide in unknown solution. First, 50.00 mL of the unknown soluton S00 M NaS20srned light ylorless was reacted with excess potassium t was filled with 0.07500 M Na S20s. The initial buret reading was turned light yellow, at n to 0.05 mL. The Is solution was then titrated with NazS,O, until the titratd which point 2 mL of starch indicator was added. The solution tu endpoint was Find titration reactions in Ch 16. turned a deep blue. A clear and colorless . Find the molarity of MnO, in the original unknown solution. Hint: reached at 42.65 ched at 42.65 mL. Find the molarity of MnO F original unknow 8. An electrochemical cell was used for a quantitative analysis of copper. A silver-silver chloride electrode was used as the anode, and a copper wire was used for the cathode. When the cell potentials for a series of Cu standards were measured, a calibration curve was obtained. The calibration equation is: Ecell 0.0128 In (1/(Cu2+D+ 0.031 0.135 grams was dissolved in 1.00 L of distilled water. The of the copper (a) (4 pts) A solid unknown sample weighing solution, when tested in the cell above, gave a reading of -0.0590 V. What is the molarity in the tested solution? (Assume that the standard solutions had concentration units of molarity.) (b) (4 points) What is the concentration of copper in the original unknown solid?Explanation / Answer
Iodometry is based on the principle that, Iodine dissolves in the iodide-containing solution to give triiodide ions, which have a dark brown color. The triiodide ion solution is then titrated against standard thiosulfate solution to give iodide again using starch indicator:
I3 + 2 S2O32 S4O62 + 3 I
From stoichiometric equation it is clear that 1 mole of triiodide reacts with 2 moles of tiosulphate .
So no.of moles of I3 formed = 1/2(no.of moles of S2O32)
n I3 formed = 1/2(nS2O32)
nS2O32 = Concentration in M x titre volume in L
Concentration = 0.07500 M
Titre Volume = (42.65 - 0.05) ml = 42.60 ml = 0.04260 L
nS2O32 = 0.07500 (mol/L) x 0.04260 L = 0.00320 mol
n I3 formed = 1/2(0.00320 moles) = 0.00160 mol
The stoichiometric equation for the reaction between KI and KMnO4 in acid mdium is,
10KI + 8H2SO4 + 2KMnO4 5I2 + 2MnSO4 + 6K2SO4 + 8H2O
Excess KI reacts with liberated iodine to form I3^- ion.
5I2 + 5 KI 5 K^+ + 5I3-
Hrere 2 moles of KMnO4 liberates 5 moles of I2 .
nKMnO4 =2/5(nI2)
Since, n I3 formed = 1/2(0.00320 moles) = 0.00160 mol
n I2 formed = 0.00160 mol
nKMnO4 =2/5(nI2) = 2/5(0.00160 mol) = 0.00064 mol
Volume of KMnO4 taken= 50.0 ml = 0.050 L
Concentration of KMnO4 = nKMnO4 / Volume of KMnO4 in L = 0.00064/0.050 = 0.01280 M
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