Question

What is the pH of a 15.9 M solution of H₂SO₄? The first proton completely dissociates; the K_a for the second proton is 1.2×10^–2.

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Question: in...typed format...

$\Rightarrow$Answer:

For the given solution of H₂SO₄ (aq) : pH = 1.92 , neglecting any negative ( - )ve value of : pH ... ( second dissociation)

$\Rightarrow$Explanation:

Following is the complete: Explanation: for the above: Answer...

• Given:

1. Molar concentration of H₂SO₄ (aq) : M_acid = 15.9 M  ( mol/L ) ...i.e. assumed = Z (mol/L )
2. Acid dissociation constant for the second proton: K_a2 = 1.2 x 10^-2
3. For the solution of H₂SO₄ (aq) : first proton: completely dissociates.

• ​​​​​​​Step - 1:

​​​​​​​For dissociation of aqueous solution of H₂SO₄ (aq) , we can form the following sequence of balanced chemical equations:

Type of dissociation	Balanced chemical equation	Identity of Equation
First proton dissociation	H2SO4 (aq)  $\rightleftharpoons$ H+(aq) + HSO4- (aq)	Equation - 1
Second proton dissociation	HSO4- (aq) $\rightleftharpoons$ H+(aq) + SO42- (aq)	Equation - 2

• Step - 2:

​​​​​​​$\Rightarrow$ Since, the dissociation of the first proton: was completed, according to: Equation -1, we can assume the concentration: of the proton , i.e. [ H⁺ ], after the first dissociation: remains constant: at:  [ H⁺ ] = Z ( mol/L )

Therefore, after dissociation of the second proton: let's assume, the concentration of : [ H⁺ ] , will become the following:

i.e. after dissociation of the second proton: $\Rightarrow$[ H⁺ ] = Z + x   ( mol/L )

$\Rightarrow$ Therefore: we will have the following molar concentrations: ( i.e. after the dissociation of second proton ) ;

1. [ H⁺ ] = Z + x   ( mol/L )
2. [HSO₄^- ] = Z - x ( mol/L )
3. [SO₄^2- ] = x   ( mol/L )

​​​​​​​Note: We have arrived at the above results, according to the following ICE Table, prepared based on: Equation - 2...

• Step - 3:

​​​​​​​Let's assume, initial concentration: of HSO₄^- (aq) = Z ( mol/L ) = 15.9 M ( given )

Therefore: following will be our ICE Table:

	[HSO4- ] ,M (mol/L)	[ H+ ], M (mol/L )	[SO42- ], M (mol/L )
Initial concentration	Z	0.0	0.0
Change ( in concentration )	Z - x	+x	+x
Equilibrium ( concentration )	Z - x	+x	+x

$\Rightarrow$Therefore:

we will have the following molar concentrations: ( i.e. after the dissociation of second proton, at Equilibrium ) ;

1. [ H⁺ ]_eq = Z + x   ( mol/L )
2. [HSO₄^- ]_eq = Z - x ( mol/L )
3. [SO₄^2- ]_eq  = x   ( mol/L )

• Step - 4:

Therefore: we will get:

$\Rightarrow$K_a2 =   ( [ H⁺ ]_eq x [SO₄^2- ]_eq ) / (  [HSO₄^- ]_eq ) =   1.2 x 10^-2   

Plugging in values, in the above: Equation: we will get the following:

$\Rightarrow$   K_a2 = 1.2 x 10^-2   = [ ( Z + x ) x ( x ) ] / ( Z - x ) ----------------------Equation - 3

$\Rightarrow$We know:  Z = 15.9 M ( mol/L )

Plugging in values in, Equation - 3, we will get the following:

• Step - 5:

​​​​​​​$\Rightarrow$x² + 15.912 x - 0.1908 = 0 -----------------Equation - 4

By solving the above: Quadratic Equation; and neglecting: (negative values of 'x' , we will get the following:

$\Rightarrow$x = 0.01198 M ( mol/L ), approx.

Therefore:

• Step - 6:

​​​​​​​Therefore: we get...

$\Rightarrow$ [ H⁺ ]_eq = Z + x = 15.9 + 0.01198 = 15.91198 M   ( mol/L )

Therefore:

$\Rightarrow$   pH = - log₁₀ [ H⁺ ]_eq =   - log₁₀ [ 15.91198 ] =  -1.20

OR, if we assume:   [ H⁺ ]_eq = x = 0.01198 M

Then, we will have the following:

'$\Rightarrow$    pH = - log₁₀ [ H⁺ ]_eq =   - log₁₀ [ 0.01198 ] =  1.92

$\Rightarrow$   pH = 1.92