Calculate the pH of a water at 25°C that contains 0.75mg/L of Carbonic acid. Assume that [H+] = [HCO3-] at equilibrium and neglect the dissociation of water. (Note: pKafor H2CO3is 6.35. Assume there is no further dissociation of HCO3-)

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gbustamantegarcia054 gbustamantegarcia054 · Answer 1 · 2019-09-09T06:00:42+02:00

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