Answer : The correct option is (d) a solution of 0.10 M NaOH
Explanation :
<u>(a) a solution of pH 3.0</u>
First we have to calculate the pOH.
![pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11](https://tex.z-dn.net/?f=pH%2BpOH%3D14%5C%5C%5C%5CpOH%3D14-pH%5C%5C%5C%5CpOH%3D14-3.0%3D11)
Now we have to calculate the
concentration.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
![11=-\log [OH^-]](https://tex.z-dn.net/?f=11%3D-%5Clog%20%5BOH%5E-%5D)
![[OH^-]=1.0\times 10^{-11}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.0%5Ctimes%2010%5E%7B-11%7DM)
Thus, the
concentration is, ![1.0\times 10^{-11}M](https://tex.z-dn.net/?f=1.0%5Ctimes%2010%5E%7B-11%7DM)
<u>(b) a solution of 0.10 M
</u>
As we know that 1 mole of
is a weak base. So, in a solution it will not dissociates completely.
So, the
concentration will be less than 0.10 M
<u>(c) a solution with a pOH of 12.</u>
We have to calculate the
concentration.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
![12=-\log [OH^-]](https://tex.z-dn.net/?f=12%3D-%5Clog%20%5BOH%5E-%5D)
![[OH^-]=1.0\times 10^{-12}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.0%5Ctimes%2010%5E%7B-12%7DM)
Thus, the
concentration is, ![1.0\times 10^{-12}M](https://tex.z-dn.net/?f=1.0%5Ctimes%2010%5E%7B-12%7DM)
<u>(d) a solution of 0.10 M NaOH</u>
As we know that
is a strong base. So, it dissociates to give
ion and
ion.
So, 0.10 M of
in a solution dissociates to give 0.10 M of
ion and 0.10 M of
ion.
Thus, the
concentration is, 0.10 M
<u>(e) a
solution of
</u>
As we know that 1 mole of
in a solution dissociates to give 1 mole of
ion and 1 mole of
ion.
So,
of
in a solution dissociates to give
of
ion and
of
ion.
The concentration of
ion is ![1\times 10^{-4}M](https://tex.z-dn.net/?f=1%5Ctimes%2010%5E%7B-4%7DM)
First we have to calculate the pH.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![pH=-\log (1.0\times 10^{-4})](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%281.0%5Ctimes%2010%5E%7B-4%7D%29)
![pH=4](https://tex.z-dn.net/?f=pH%3D4)
Now we have to calculate the pOH.
![pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10](https://tex.z-dn.net/?f=pH%2BpOH%3D14%5C%5C%5C%5CpOH%3D14-pH%5C%5C%5C%5CpOH%3D14-4%3D10)
Now we have to calculate the
concentration.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
![10=-\log [OH^-]](https://tex.z-dn.net/?f=10%3D-%5Clog%20%5BOH%5E-%5D)
![[OH^-]=1.0\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.0%5Ctimes%2010%5E%7B-10%7DM)
Thus, the
concentration is, ![1.0\times 10^{-10}M](https://tex.z-dn.net/?f=1.0%5Ctimes%2010%5E%7B-10%7DM)
From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.
Hence, the correct option is (d)