2.68×10^-3 moles of a solution containing an ion A^n acquire 1.61×10^...
2.68×10^-3 moles of a solution containing an ion A^n acquire 1.61×10^...
To determine the value of n in the equation A^n + MnO4^- -> AO3^- in acidic medium, we need to use the stoichiometry of the reaction.
Let's start by determining the number of moles of MnO4^- required to oxidize A^n to AO3^- based on the given information:
Moles of MnO4^- = 1.61×10^-3 moles
Next, we need to find the molar ratio between MnO4^- and A^n. This can be determined by balancing the oxidation half-reaction, which involves the transfer of electrons:
MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O
From this balanced equation, we can see that 1 mole of MnO4^- requires 5 moles of electrons. Therefore, the molar ratio between MnO4^- and A^n is 5:1.
Now, we can calculate the number of moles of A^n by multiplying the moles of MnO4^- by the molar ratio:
Moles of A^n = (1.61×10^-3 moles MnO4^-) * (1 mole A^n / 5 moles MnO4^-)
= 3.22×10^-4 moles A^n
Finally, we can calculate the value of n by comparing the moles of A^n with the given amount of A^n in the solution:
n = (moles of A^n) / (moles of solution)
= (3.22×10^-4 moles) / (2.68×10^-3 moles)
≈ 0.12
Since n represents a whole number, we can round 0.12 to the nearest whole number, which is 0. Therefore, the value of n is 0.
Answer: The value of n is 0.