Given :
0.00072 M solution of 
at 
.
To Find :
The concentration of 
and pOH .
Solution :
1 mole of gives 2 moles of ions .
So , 0.00072 M mole of gives :
![[OH^-]=2 \times 0.00072\ M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D2%20%5Ctimes%200.00072%5C%20M)
![[OH^-]=0.00144\ M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.00144%5C%20M)
![[OH^-]=1.44\times 10^{-3}\ M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.44%5Ctimes%2010%5E%7B-3%7D%5C%20M)
Now , pOH is given by :
![pOH=-log[OH^-]\\\\pOH=-log[1.44\times 10^{-3}]\\\\pOH=2.84](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D-log%5B1.44%5Ctimes%2010%5E%7B-3%7D%5D%5C%5C%5C%5CpOH%3D2.84)
Hence , this is the required solution .
Answer:
c
Explanation:
I think that's the answer
A. Archilles ! Hope this helped :)!
Write a balance equation for the reaction between the analyte and the titrant.
Calculate the # of moles of titrant using the volume of titrant required and the concentration of titrant.
Calculate the # of moles of analyte using the stoichiometric coefficients of the equation.
Calculate the concentration of the analyte using the number or moles of analyte and the volume of analyte titrated.
