Answer : The concentration of 
ion is, 
Solution : Given,
pH = 4.20
First we have to calculate the pOH.
As we know that,
Now we have to calculate the concentration of ion.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
![9.8=-\log [OH^-]](https://tex.z-dn.net/?f=9.8%3D-%5Clog%20%5BOH%5E-%5D)
![[OH^-]=1.58\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.58%5Ctimes%2010%5E%7B-10%7DM)
Therefore, the concentration of ion is,
Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
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