In the spectrophotometric determination of manganese in steel, a 0.250 gram samp

Question

In the spectrophotometric determination of manganese in steel, a 0.250 gram sample of unknown steel is dissolved and the manganese oxidized to MnO_4^-. This solution is then diluted to 500 ml in a volumetric flask. The quantity 0.600 grams of a steel sample, certified to contain 0.25% Mn, is treated in exactly the same way. The Spectronic 20, set at 540 nm to match the absorption band of the MnO_4^- ion, is adjusted so that the instrument reads 0% transmission with no sample in the Spectronic 20 and 100% transmission with a distilled water blank in the instrument. The % transmission unknown solution (%T(x)) is found to be 58.0 %, while the % transmission of the known solution (%T(k)) is found 39.0%. Determine mg Mn(x), and the percentage of Mn in the 0.250 gram sample of unknown steel. mg Mn(x) = mg Mn(k) times [2 - log%T(x))/times (2 - log%T(k))] Theory of Neutralization Titrations

Explanation / Answer

The theory used is pretty simple here.

Putting in the %T values for the known and unknown solutions, we get:

mg Mn(x) = mg Mn(k) * ( (2 - log(58))/(2 - log(39)) ) = 0.578

Also, 0.60g of the known sample is known to have 0.25% Mn

mg Mn(k) = (0.25*0.60)/100 = 0.0015 g = 1.5 mg

So,

mg Mn(x) = 0.578*1.5 = 0.867 mg

% Mn in the sample = (0.867*10-3)/(0.25)*100 = 0.3468%

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