Question

What mass of potassium hypochlorite (FW-90.6 g/mol) must be added to 4.50 x 10 mL of water to give a solution with pH 10.20? [Ka(HCIO) 4.0 x 10-8] 0.032g ? 2.4 g 04.1 g 9.1 g 20. g

Answer 1

Answer : The mass of potassium hypochlorite is, 4.1 grams.

Explanation : Given,

pH = 10.20

Volume of water = $4.50\times 10^2ml=0.45L$

The decomposition of KClO  will be :

$KClO\rightarrow K^++ClO^-$

Now the further reaction with water $(H_2O)$ to give,

$ClO^-+H_2O\rightarrow HClO+OH^-$

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-10.20=3.8$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$3.8=-\log [OH^-]$

$[OH^-]=1.58\times 10^{-4}M$

Now we have to calculate the base dissociation constant.

Formula used : $K_b=\frac{K_w}{K_a}$

Now put all the given values in this formula, we get :

$K_b=\frac{1.0\times 10^{-14}}{4.0\times 10^{-8}}=2.5\times 10^{-7}$

Now we have to calculate the concentration of $ClO^-$.

The equilibrium constant expression of the reaction  is:

$K_b=\frac{[OH^-][HClO]}{[ClO^-]}$

As we know that, $[OH^-]=[HClO]=1.58\times 10^{-4}M$

$2.5\times 10^{-7}=\frac{(1.58\times 10^{-4})^2}{[ClO^-]}$

$[ClO^-]=0.0999M$

Now we have to calculate the moles of $ClO^-$.

$\text{Moles of }ClO^-=\text{Molarity of }ClO^-\times \text{Volume of solution}$

$\text{Moles of }ClO^-=0.0999mole/L\times 0.45L=0.0449mole$

As we know that, the number of moles of $ClO^-$ 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.

$\text{Mass of }KClO=\text{Moles of }KClO\times \text{Molar mass of }KClO$

$\text{Mass of }KClO=0.0449mole\times 90.6g/mole=4.07g\approx 4.1g$

Therefore, the mass of potassium hypochlorite is, 4.1 grams.