Respuesta :
Answer : The mass of potassium hypochlorite is, 4.1 grams.
Explanation : Given,
pH = 10.20
Volume of water = [tex]4.50\times 10^2ml=0.45L[/tex]
The decomposition of KClO will be :
[tex]KClO\rightarrow K^++ClO^-[/tex]
Now the further reaction with water [tex](H_2O)[/tex] to give,
[tex]ClO^-+H_2O\rightarrow HClO+OH^-[/tex]
First we have to calculate the pOH.
[tex]pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-10.20=3.8[/tex]
Now we have to calculate the [tex]OH^-[/tex] concentration.
[tex]pOH=-\log [OH^-][/tex]
[tex]3.8=-\log [OH^-][/tex]
[tex][OH^-]=1.58\times 10^{-4}M[/tex]
Now we have to calculate the base dissociation constant.
Formula used : [tex]K_b=\frac{K_w}{K_a}[/tex]
Now put all the given values in this formula, we get :
[tex]K_b=\frac{1.0\times 10^{-14}}{4.0\times 10^{-8}}=2.5\times 10^{-7}[/tex]
Now we have to calculate the concentration of [tex]ClO^-[/tex].
The equilibrium constant expression of the reaction is:
[tex]K_b=\frac{[OH^-][HClO]}{[ClO^-]}[/tex]
As we know that, [tex][OH^-]=[HClO]=1.58\times 10^{-4}M[/tex]
[tex]2.5\times 10^{-7}=\frac{(1.58\times 10^{-4})^2}{[ClO^-]}[/tex]
[tex][ClO^-]=0.0999M[/tex]
Now we have to calculate the moles of [tex]ClO^-[/tex].
[tex]\text{Moles of }ClO^-=\text{Molarity of }ClO^-\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }ClO^-=0.0999mole/L\times 0.45L=0.0449mole[/tex]
As we know that, the number of moles of [tex]ClO^-[/tex] are equal to the number of moles of KClO.
So, the number of moles of KClO = 0.0449 mole
Now we have to calculate the mass of KClO.
[tex]\text{Mass of }KClO=\text{Moles of }KClO\times \text{Molar mass of }KClO[/tex]
[tex]\text{Mass of }KClO=0.0449mole\times 90.6g/mole=4.07g\approx 4.1g[/tex]
Therefore, the mass of potassium hypochlorite is, 4.1 grams.
