A carbonic anhydrase generates H+ at a rate constant (K3) of 10^(6) per second.
ID: 894707 • Letter: A
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)?
**CAN YOU BREAK DOWN THE SUBSTITUTION PART IN QUESTION 2, I DON'T UNDERSTAND HOW [A] IS 9.09 AND HOW [HA]= 91, ITS THE ACTUAL MATH THAT I DON'T GET.
Explanation / Answer
1) H+ is formed at a rate of Kcat is 106 /mol /sec.
In one min, 60(106) = 60000000 H+ /mol /min.
60000000(1E-9M Carbonic Anhydrase) = .006M H+ formed.
pH = -log[H+] = -log[.006]
pH = 2.22
2)
100 = [ A ]+ [AH], where Bt stands for total buffer
pH = 7, pKa = 8
pH = pKa + log( [ A ]/ [AH]
7 = 8 + log([ A ]/ [AH]
-1 = log(/[ A ]/ [AH]
[HA] = 90.9mM, [A]= 9.09mM -------------------------------> your doubt
pH = pKa + log(/[ A ]/ [AH]
([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
in your doubt total buffer concentration = salt + acid = 100mM
[A] = salt
[HA]= acid
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