Question

1)Hypochlorous acid (HOCl), a weak acid (pKa = 7.6) can be used for disinfecting water. If 1 mg of HOCl is added to pure water to make up 1 L of liquid volume, what is the equilibrium pH?

(b) The pH of the solution in part (a) is to be raised to 7.0 by addition of NaOH, a strong base. What mass (in grams) of NaOH must be added?

Answer 1

1)

The reaction of HOCl with water is as follows

HOCl + H2O <------------> H3O+ + OCl-

H3O+ formed at equilibrium is responsible for the pH of the solution.

Let's find molarity of the solution now

molarity = moles of HOCl/ Liters of solution

Moles of HOCl = 1 mg * 1 g/1000 mg * 1 mol / 52.46 g = 1.906 x 10^-5 mol

Molarity = 1.906 x 10^-5 / 1 L = 1.906 x 10^-5 M

Let's draw ICE table for above equation

HOCl + H2O <--------------> H3O+ + OCl-

	HOCl	H2O	H3O+	OCl-
I	1.906 x 10^-5	-	0	0
C	-x	-	+x	+x
E	1.906 x 10^-5 - x	-	x	x

Equilibrium constant for above reaction is written as

Ka = [H3O+] [ OCl-] / HOCl]   

pKa is given as 7.6

pKa = - log [ Ka]

7.6 = -log [Ka] ...............Divide both sides by -1

-7.6 = log [Ka] ...............Take inverse function of log on both sides which is 10^ to cancel out log

10^-7.6 = Ka

Ka = 2.511 x 10^-8

Substituting this value of Ka and values from ICE table in above equation we get

2.511 x 10^-8 = (x) (x) / 1.906 x 10^-5 - x ............. we can ignore x on the bottom as Ka is very small

2.511 x 10^-8 = x^2 / 1.906 x 10^-5

x^2 = 4.786 x 10^-13

x = 6.918 x 10^-7 M

so we get [H3O+] = 6.736 x 10^-7 M

pH = - log [H3O+]

pH = -log ( 6.736 x 10^-7)

pH = 6.16

The equilibrium pH is 6.16

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b.

Let us take a look at the reaction that takes place when HOCl reacts with NaOH

HOCl + NaOH <------------> NaOCl + H2O

From the above equation we can see that at equilibrium we get HOCl which is a weak acid and its strong salt NaOCl

This system represents a buffer and therefore we can use Henderson - Hasselbalch equation to find pH of this system

The equation is written as follows

pH = pKa + log [ base] / [acid]

They want pH to be raised to 7, so here we have pH = 7

pKa is given and its 7.6

We calculated concentration of acid in part a which is 1.906 x 10^-5 M

We are finding here concentration of conjugate base of HOCl

Substituting above values in the equation we get

7 = 7.6 + log [ base] / [ 1.906 x 10^-5 ]

-0.6 = log  [ base] / [ 1.906 x 10^-5 ] ....... Take 10^on both sides

10^-0.6 =   [ base] / [ 1.906 x 10^-5 ]

0.251 =  [ base] / [ 1.906 x 10^-5 ]

[base] = 4.788 x 10^-6 M

concentration of NaOCl is 4.788 x 10^-6 M

But if we look at the above reaction, we can see that NaOCl depends upon NaOH that is being added

Therefore concentration of NaOH is same as that of NaOCl

concentration of NaOH = 4.788 x 10^-6 M

moles of NaOH = 4.788 x 10^-6 mol/L * 1 L ......... We have used 1 L of solution

moles of NaOH = 4.788 x 10^-6

Mass of NaOH used = mol * molar mass

= 4.788 x 10^-6 mol * 39.997 g/ mol

= 1.915 x 10^-4 g

1.915 x 10^-4 g NaOH must be added to raise the pH to 7