Answer:
pH = 5.675
Explanation:
H2CO3 ↔ H3O+ + HCO3-
∴ C H2CO3 = 0.75 mg/L * ( g / 1000 mg ) * ( mol / 62.03 g ) = 1.209 E-5 M
∴ Ka = ( [ HCO3- ] * [ H3O+ ] ) / [ H2CO3 ]
∴ pKa = 6.35 = - Log ka
⇒ Ka = 4.467 E-7 = ( [ HCO3- ] * [ H3O+ ] ) / [ H2CO3 ]
assuming that the dissociation of HCO3- is not signicant, we have:
mass balance:
C H2CO3 = [ H2CO3 ] + [ HCO3- ] = 1.209 E-5
charge balance:
⇒[ H3O+ ] = [ HCO3- ]............. assumption given for equilibrium
⇒ 1.209 E-5 - [ H3O+ ] = [ H2CO3 ]
replacing in Ka:
⇒ 4.467 E-7 * ( 1.209 E-5 - [ H3O+ ] ) = [ H3O+ ]²
⇒ 5.401 E-12 - 4.467 E-7 [ H3O+ ] = [ H3O+ ]²
⇒ [ H3O+ ]² + 4.467 E-7 [ H3O+ ] - 5.401 E-12 = 0
⇒ [ H3O+ ] = 2.111 E-6 M
⇒ pH = - Log [ H3O+]
⇒ pH = 5.675
