A carbonic anhydrase generates H+ at a rate constant (K3) of 10^(6) per second.

Question

A carbonic anhydrase generates H+ at a rate constant (K3) of 10^(6) per second.  Assume the rate constant of the enzyme is not affected by pH, and the initial pH of the reaction solution = 7.0.  


1) What will be the pH in the reaction solution after 0.1 nM carbonic anhydrase was incubated with unlimited substrate (CO2) for 1 min?



2) What will be the pH in the above reaction solution (0.1 nM carbonic anhydrase incubated with unlimited substrate (CO2) for 1 min) if 100 mM Tris (pKa = 8.0) was included as buffer (initial pH = 7.0)?

Explanation / Answer

1) H+ is formed at a rate of Kcat is 10 ^ 6 /mol /sec.

In one min, 60(10^6) = 60000000 H+ /mol /min.

60000000(1E-9M Carbonic Anhydrase) = .006M H+ formed.

pH = -log[H+] = -log[.006] = 2.22



2) [Bt] = [B] + [BH+], where Bt stands for total buffer, [Bt] = 100mM

pH = 7, pKa = 8

pH = pKa + log([B]/[BH+])

7 = 8 + log([B]/[BH+])

-1 = log([B]/[BH+])

1/10 = [B]/[BH+], (1/10)[BH+] = [B]

(make substitution)
at pH = 7, [Bt] = 1/10[BH+] + [BH+]

[BH+] = 90.9mM, [B] = 9.09mM

pH = pKa + log([B]/[BH+])

([H+] changes buffer ratio)
pH = 8 + log[(9.09E-3 - .006)/(90.9 +.006)]

pH = 8 + log(.0031/.969)

pH = 8 + (-1.5)

pH = 6.5

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