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pts) A 25.0 mL sample of water from the lake was acidified and treated with excess sodium molybdate, leading to the formation of 12 -molybdophosphoric acid (12-MPA):
+ 24 H+ =====-> H3PMO12040
+ 24Nat +
12 H20
HsPO4 + 12 NazMoO4
The 12-MPA was separated from the excess sodium molybdate by extraction with two successive 10.0 mL portions of iso-amyl alcohol. When the combined extract was aspirated into the flame of an atomic absorption spectrometer set-up for measuring molybdenum, an absorbance of 0.662 was obtained. The absorbance of a 5.00- ppm Mo standard in the same volume of iso-amyl alcohol was 0.203.Calculate the ppm PO43 in the sample.

Sagot :

Zueyae2006idn

Answer:

therefore, the concentration of 12-MPA in the sample is 1.81 x 10^-5 ppm.

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