Question

Determine the pH of a solution that is 0.00596 M HCl and 0.0522 M HClO2. The ?a of HClO2 is 1.1×10−2 .

Answer 1

Sol: Information Provided in the question:

?a of HClO2 is = 1.1×10⁻²

Concentration of HCl =  0.00596 M

Concentration of HClO2 =  0.0522 M

Step (1): To determine ph of a solution that is 0.00596 M HCl :

Ionization Reaction for HCl can takes place like :

Concentration of Component	HCl	H+	Cl-
Initial	0.00596	0	0
Change	- r	+r	+r
Final	(0.00596 - r)	+r	+r

So final EquilibriumExpression:

Ka = ( [H⁺] [Cl^-] ) / [HCl]

1.1×10⁻² = ( [+r] [+r] ) / [(0.00596 - r)]

Then

1.1×10⁻² x(0.00596 - r) = r x r

6.556x10^-5 - 1.1×10⁻² x r = r²

r² + 1.1×10⁻² x r = 6.556x10^-5

On solving the quadratic equation:

r = + 4.28825x10^-3; - 0.01528825 ;

Hence Positive Value is taken :

r = + 4.28825x10^-3 = [H⁺]

Then the equation of Ph:

ph = - log [+ 4.28825x10^-3] = 2.3676642

Hence ph of a solution that is 0.00596 M HCl = 2.3676642

Step (2): To determine ph of a solution that is 0.0522 M HClO₂:

Ionization Reaction for HClO₂ can takes place like :

Concentration of Component	HClO2	H+	ClO2-
Initial	0.0522	0	0
Change	- r	+r	+r
Final	(0.0522 - r)	+r	+r

So final Equilibrium Expression:

Ka = ( [H⁺] [ClO₂^-] ) / [HClO₂]

1.1×10⁻² = ( [+r] [+r] ) / [(0.0522 - r)]

1.1×10⁻² x (0.0522 - r) = r x r

5.742x10^-4 -1.1×10⁻² x r = r²

r² + 1.1×10⁻² x r = 5.742x10^-4

On solving the quadratic equation:

r = + 0.0190855 ; -0.030085

Hence Positive Value is taken :

r = + 0.0190855 = [H⁺]

Then the equation of Ph:

ph = - log [+ 0.0190855] = 1.719296

Hence ph of a solution that is 0.0522 M HClO₂: = ph = - log [+ 0.0190855] = 1.719296