What is the pH of a 15.9 M solution of H2SO4? The first proton completely dissociates; the Ka for the second proton is 1.2×10–2.
Following is the - complete Answer -&- Explanation: for the given: Question: in...typed format...
Answer:
For the given solution of H2SO4 (aq) : pH = 1.92 , neglecting any negative ( - )ve value of : pH ... ( second dissociation)
Explanation:
Following is the complete: Explanation: for the above: Answer...
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) H+(aq) + HSO4- (aq) | Equation - 1 |
Second proton dissociation | HSO4- (aq) H+(aq) + SO42- (aq) | Equation - 2 |
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: [ H+ ] = Z + x ( mol/L )
Therefore: we will have the following molar concentrations: ( i.e. after the dissociation of second proton ) ;
Note: We have arrived at the above results, according to the following ICE Table, prepared based on: Equation - 2...
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 |
Therefore:
we will have the following molar concentrations: ( i.e. after the dissociation of second proton, at Equilibrium ) ;
Therefore: we will get:
Ka2 = ( [ H+ ]eq x [SO42- ]eq ) / ( [HSO4- ]eq ) = 1.2 x 10-2
Plugging in values, in the above: Equation: we will get the following:
Ka2 = 1.2 x 10-2 = [ ( Z + x ) x ( x ) ] / ( Z - x ) ----------------------Equation - 3
We know: Z = 15.9 M ( mol/L )
Plugging in values in, Equation - 3, we will get the following:
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:
x = 0.01198 M ( mol/L ), approx.
Therefore:
Therefore: we get...
[ H+ ]eq = Z + x = 15.9 + 0.01198 = 15.91198 M ( mol/L )
Therefore:
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:
' pH = - log10 [ H+ ]eq = - log10 [ 0.01198 ] = 1.92
pH = 1.92
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