Manganese is frequently found in hard water samples. A container of hard water w

Question

Manganese is frequently found in hard water samples. A container of hard water was evaporated leaving 0.10 g of solid residue. This residue was dissolved in 10 ml of chloroform. 3, 5-Dibromo-2, 4-dihydroxy-acetophenone oxime (DDAO) is known to complex Min(II) in a 1:2 ratio of metal ion to ligand. A 1 times 10^-2 M solution of DDAO is added to 10 ml of the chloroform solution. Measurement of absorbance at 395 nm produced the following results: How many milligrams of manganese were present in the hard water residue?

Explanation / Answer

We have to use the method of standard addition and extrapolate to find the concentration of Mn (II) in the original sample matrix.

Plot the absorbance vs volume of DDAO (mL) as below:

The graph displays a curved patter, however, the initial part of the curve (marked with yellow points) is more or less linear and we use that portion to extrapolate to zero. The equation for the linear portion of the curve is shown alongside the curve.

Extrapolation to zero means that we will find the concentration of the sample responsible for absorption of light when the absorbance is zero.

Put y = 0 in the equation and ignore negative sign and write

0 = 0.0561x + 0.1137

===> x = 0.1137/0.0561 = 2.027 2.0

The concentration of the Mn(II) sample is 2.0*1.0*10-2 M (note that the concentration of DDAO was 1.0*10-2 M) = 2.0*10-2 M

Since Mn(II) and the complex have a ratio of 1:1 (1 mole of Mn forms 1 mole of the complex), we must have, the concentration of Mn(II) = 2.0*10-2 M.

Volume of solution = 10 mL = (10 mL)*(1 L/1000 mL) = 0.01 L.

Moles of Mn = (2.0*10-2 mol/L)*(0.01 L) = 2.0*10-4 mole.

Molar mass of Mn = 54.94 g mol-1; therefore mass of Mn in 2.0*10-4 mole Mn = (54.94 g mol-1)*(2.0*10-4 mole) = 0.010988 g 0.011 g = (0.011 g)*(1000 mg/1 g) = 11 mg (ans).

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