The ocean has natural carbonate buffer system, which helps to regulate the pH of

Question

The ocean has natural carbonate buffer system, which helps to regulate the pH of sea water. Write the two successive dissociation reactions for carbonic acid (H_2CO_3), and their associated acid dissociation constants (K_a expressions). What is the pH of a buffer that is created by combining 65.0mL of 1.00 M H_2CO_3 and 35.0 mL of 0.800 M NaHCO_3? Assume the second dissociation is negligible and don't forget to account for dilution. What would the pH of the solution be after 1.5mL of 2.5 M HCl is added?

Explanation / Answer

a) H2CO3(aq) <----> HCO3-(aq) + H+ (aq)

Ka1 = [HCO3-][H+] / [H2CO3]

HCO3-(aq) <---> CO3 ^2-(aq) + H+ (aq)

Ka2 = [CO3^2-] [H+] /[HCO3-]

b) for buffer we have Henderson eq

pH = pka1 + log [HCO3-]/[H2CO3]

final volume after mixing H2cO3 , NaHCO3 = 65+35 = 100 ml

we use dilution formula M1V1 = M2V2 to find final Molarities of H2CO3 and NaHCO3

for H2CO3   1 x 65 = M2 x 100 , M2 = 0.65 M is final Molarity

for NaHCO3   0.8 x 35 = M2 x 100 , final Molarity fo NaHCO3 = 0.28 M

now pH = 3.6 + log ( 0.28/0.65)            ( pka1= 3.6 taken from web)

      = 3.234

c) H+ moles = HCl moles added = M x V (in L) = 2.5 x ( 1.5/1000) = 0.00375

H+ reacts with HCO3- to form H2CO3

initial H2CO3 moles = M x V = 0.65 x 100/1000 = 0.065

initial HCO3- moles = 0.28 x 100/1000 = 0.028

now after H+ addition HCO3- moles = 0.028-0.00375 = 0.02425

H2CO3 moles now = 0.065+0.00375 = 0.06875

total vol now = 100+1.5 = 101.5 ml

[HCO3-] = 0.02425/0.1015 = 0.2389

[H2CO3] = 0.06875/0.1015= 0.67734

pH = 3.6 + log ( 0.2389/0.6774)

= 3.15

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