Wheeling Jesuit University- CHE 316-Spring18 S ng Jump to.. Sapling Learning 23.

Question

Wheeling Jesuit University- CHE 316-Spring18 S ng Jump to.. Sapling Learning 23.92 ml of a Pb solution, containing excess Pb, was added to 12.25 mL of a 2.3-dimercapto-1- propanol (BAL) solution of unknown concentration, forming the 1:1 Pb-BAL complex. The excess Pt was titrated with 0 0106 M EDTA, requiring 9.71 mL to reach the equivalence point. Separately, 41.75 ml of the EDTA solution was required to titrate 29 25 ml of the Pb solution. Calculate the BAL concentration, in molarity, of the original 1225-mL solution. OPrevous (> Give up & View Solden Check Aweer @N". Ert Pictu 7 2 3 4

Explanation / Answer

Let´s start with the back titration The excess Pb2+ required 9.71 mL of 0.0106 M = 0.0106 mol/L EDTA.

1 mole of Pb needs 1 mole of EDTA, so let´s find out the moles of EDTA needed

0.00971 * 0.0106 = 0.000102926 moles of EDTA = moles of Pb

Let´s find the molar concentration of Pb from the second titration

we have that, 41.75 mL of 0.0106 M EDTA was required to titrate 29.5 mL of Pb2+ solution as per the above equation.

let´s use the next equation since moles must be the same

Molarity 1 * Volume 1 = Molarity 2 * Volume 2, we want to find molarity 2

41.75 * 0.0106 = 29.25 * M2

M2 = 41.75* 0.0106 / 29.25 =0.01513 M, this is the molarity of Pb in the solution, we can calculate the number of moles in 23.92 ml (23.92 ml = 0.02392 L, just divide by 1000)

moles = Molarity * Volume = 0.01513 * 0.02392 = 0.0003619 moles of original Pb

we know that there was an original excess so

moles that reacted with Bal are = original moles of Pb - moles reacted in the back titration

moles of Bal that reacted = 0.0003619 -  0.000102926= 0.0002589 moles of Bal that reacted, remember that this is a 1 to 1 reaction

The BAL solution originally have 12.25 ml,

12.25 ml * 1 L / 1000 ml = 0.01225 L

Molarity = 0.0002589 / 0.01225 = 0.02113 M , this is the molarity of BAL in the original solution of 12.25 ml

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