Respuesta :
Answer:
Explanation:
H₂A ⇄ HA⁺ + A⁻
K₁ = [ HA⁺] [A⁻ ] / [ H₂A ]
HA⁺ ⇄ H⁺ + A⁻
K₂ = [ H⁺] [ A ⁻ ] / [ HA⁺ ]
H₂A ⇄ HA⁺ + A⁻
.100 0 0
.100 - C C C
K₁ = [ HA⁺] [A⁻ ] / [ H₂A ]
Putting the values
1.00 x 10⁻⁴ = C X C / .1 - C
10⁻⁵ - 10⁻⁴ C = C²
C² + 10⁻⁴ C - 10⁵ = 0
C = 8 X 10⁻² M
So concentration of H₂A that is [ H₂A ] = 1 - C = 0 .1 - 8 X 10⁻² M
= .02 = 2 x 10⁻² M
NaHA ⇄ Na⁺ + HA⁻
NaHA is a strong acidic salt so it will ionise 100 %
NaHA ⇄ Na⁺ + HA⁻
.1 .1 .1
concentration of HA⁻ = .1 M
Na₂A ⇄ 2Na⁺ + A⁻²
Na₂A is also a strongly ionic salt so it will dissociate 100 % .
Na₂A ⇄ 2Na⁺ + A⁻²
.1 .1 .1
concentration of A⁻² = .1 M
HA⁺ ⇄ H⁺ + A⁻
8 X 10⁻² C₁ C₁
K₂ = [ H⁺] [ A ⁻ ] / [ HA⁺ ]
Putting the values
1.00 x 10⁻⁸ = C₁ X C₁ / C
1.00 x 10⁻⁸ = C₁² / 8 X 10⁻²
C₁² = 8 x 10⁻¹⁰
C₁ = 2.828 x 10⁻¹⁰
[ H⁺] = 2.828 x 10⁻¹⁰
pH = - log [ H⁺] = - log 2.828 x 10⁻¹⁰
= 10 - log 2.828
= 10 - .45
= 9.55 .
