Can you help me with A-chem question? (CH 16) A 50.00-mL solution containing an

Question

Can you help me with A-chem question? (CH 16)

A 50.00-mL solution containing an unknown amount of UO_4^2^- was passed through a Jones reductor, converting UO_4^2- to U^3+. Upon exposure to atmospheric oxygen, the U^3+ was reoxidized to U^4+. The solution was then diluted to 100.0 ml. using 1 M H_2SO_4 and titrated with 15.23 mL of 0.0954 M KMnO_4 to reach the purple endpoint. A blank solution subjected to the same procedure required 0.14 mL. Balance the titration reaction and determine the concentration of UO_4^2- in the original solution. MnO_4^- +U^4= rightarrow Mn^2+ + UO_2^2+

Explanation / Answer

Here

MnO4 -    ---- > Mn2+

+7         +2 by 5 e-

U^4+    - ---àUO2 ^2+

+4 ------à +6 by releasing 2    e-

2*[MnO4 -    + 5 e-     ---- > Mn2+]

5*[U^4+    - ---àUO2 ^2+    + 2e- ]

2 MnO4 - +5 U^4+       ---- > 2 Mn2+ 5 UO2 ^2+

Here the volume of KMnO4 which is actual used = 15.23-0.14 ml

= 15.09 ml

= 0.01509 L

Molarity of KMno4 = 0.0954 M

Number of moles = molarity * volume in L

= 0.0954 * 0.01509 L

= 0.00144 Moles KMnO4 or 0.00144 Moles MnO4 -

Now calculate the moils of U^4+       

0.00144 Moles MnO4 - *5.00 moles U^4+ /2 .00 Moles MnO4 -

=0.0036 moles of U^4+ this is equal to the moles of UO2^2+

Volume = 100 ml = 0.100 L

      

Molarity = 0.0036 moles /0.100 L

= 0.036 M

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---