Question

A solution of household bleach contains 5.25% sodium hypochlorite, NaOCl, by mass. Assuming that the density of bleach is the same as water, calculate the volume of household bleach that should be diluted with water to make 500.0 mL of a pH = 10.00 solution. Use the K a of hypochlorous acid found in the chempendix.

Answer 1

Answer :- 28.34 ml

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Explanation :-

Given :-

5.25 % NaOCl solution.

density of NaOCl solution = 1 g/ml

volume of diluted NaOCl solution = 500 ml

pH = 10

we have,

Ka of HClO =

Molar mass of NaOCl = 74.4 g/mol

First we calculate the concentration of stock NaOCl solution.

5.25 % NaOCl means 5.25 g NaOCl dissolved in 94.75 g water making 100 g NaOCl solution

We know that,

therefore, volume of stock NaOCl solution is

Now calculate moles of NaOCl

Formula :-

Therefore moles of NaOCl is

Now calculate concentration of stock NaOCl solution

Formula :-

therefore,

Now NaOCl is a salt it ionisation as

The ClO- ion reacts with water as

..(A)

For reaction (A) the Kb expression is

....(1)

We know

we know and  

therefore,

Let, C = concentration of ClO- (i.e. NaOCl) after dilution

We construct an ICE table for reaction (A)

	ClO-	+	H2O		HClO	+	OH-
initial	C		-		0		0
change	- x		-		+x		+x
equilibrium	C-x		-		x		x

Thus equation (1) becomes

  

i.e. .....(2)

We have, pH = 10

we know that, pOH = 14 - pH

therefore, pOH = 14 -10 = 4

Again we know that,

i.e.

therefore,  

Hence, equation (2) becomes

  

i.e.

i.e.

  

i.e.

since, 0.04 >>>> 0.0001

hence, (0.04 - 0.0001) = 0.04 M

Now we calculate volume of NaOCl diluted

we know ,  

we have,

M1 = molarity of stock NaOCl = 0.7057 M

V1 = volume of stock NaOCl = ?

M2 = molarity of diluted NaOCl = 0.04 M

V2 = volume of diluted NaOCl = 500 ml

therefore,

  

i.e.

therefore,

the volume of household bleach that should be diluted with water to make 500.0 mL of a pH = 10.00 solution will be 28.34 ml

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