Biochem question; Can someone explain protonated:deprotonated calculation. If pk
ID: 317569 • Letter: B
Question
Biochem question; Can someone explain protonated:deprotonated calculation. If pka = 4.25 and the pH is 4. What is the ratio of protonated:depritonated forms of this peptide?I get that pH-pKa = log(A-)/(HA) I get an answer of 0.56 but I am not sure what this represents... is 0.56 the protonated or deprotonated form?
My professor additionally had the equation as pH =pKa+log (HA)/(A-); why is the ratio flipped?
Biochem question; Can someone explain protonated:deprotonated calculation. If pka = 4.25 and the pH is 4. What is the ratio of protonated:depritonated forms of this peptide?
I get that pH-pKa = log(A-)/(HA) I get an answer of 0.56 but I am not sure what this represents... is 0.56 the protonated or deprotonated form?
My professor additionally had the equation as pH =pKa+log (HA)/(A-); why is the ratio flipped?
Biochem question; Can someone explain protonated:deprotonated calculation. If pka = 4.25 and the pH is 4. What is the ratio of protonated:depritonated forms of this peptide?
I get that pH-pKa = log(A-)/(HA) I get an answer of 0.56 but I am not sure what this represents... is 0.56 the protonated or deprotonated form?
My professor additionally had the equation as pH =pKa+log (HA)/(A-); why is the ratio flipped?
I get that pH-pKa = log(A-)/(HA) I get an answer of 0.56 but I am not sure what this represents... is 0.56 the protonated or deprotonated form?
My professor additionally had the equation as pH =pKa+log (HA)/(A-); why is the ratio flipped?
Explanation / Answer
Henderson Hasselbalch equation is PH = Pka+ log(A-)/(HA) = PH = Pka+ log(conjugated base)/(Acid)
PH-Pka= log(A-)/(HA)
log(A-)/(HA) = PH-Pka
Put the value of PH & Pka
log(A-)/(HA) = 4 - 4.25
log(A-)/(HA) = -0.25
We have to calculate the anti log of -0.25
(A-)/(HA) = 10-0.25 = 0.5623
The given equation by your professor is not correct equation. By inverting the term log(A-)/(HA) we will get the correct equation PH = Pka- log(HA)/(A-)
log(HA)/(A-) = Pka- PH
By putting the values log(HA)/(A-) = Pka- PHg(HA)/(A-) = 4.25- 4
log(HA)/(A-) = 0.25
by taking anti log (HA)/(A-) = 100.25 = 1.778
(HA)/(A-) ratio will be the protonated: deprotonated form, so answer will be 1.778
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