the spectrophotometric determination of manganese in steel, a 0.250 gram sample

Question

the spectrophotometric determination of manganese in steel, a 0.250 gram sample of unknown steel is dissolved manganese oxidized to MnO_4^-. This solution is then diluted to 500 ml in a volumetric flask. The quantity 0.60 of a steel sample, certified to contain 0.25% Mn, is treated in exactly the same way. The Spectronic 20, set a match the absorption band of the MnO_4^- ion, is adjusted so that the instrument reads 0% transmission with no; in the Spectronic 20 and 100% transmission with a distilled water blank in the instrument. The % transmission known solution (%T(x)) is found to be 58.0 %, while the % transmission of the known solution (%T(k)) is for)%. ine 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 pic is not uploaded completely, and the data is missing on the sides of the pic.

The theory used, howeever 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(60)) ) = 1.066

Here I have assumed %T(k) as 60% because the data is not shown in the pic.

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) = 1.066*1.5 = 1.599 mg

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.