Biochem question Can someone explain protonateddeprotonated

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?

Henderson Hasselbalch equation is P^H = P^ka+ log(A^-)/(HA) = P^H = P^ka+ log(conjugated base)/(Acid)

P^H-P^ka= log(A-)/(HA)

log(A-)/(HA) = P^H-P^ka

Put the value of P^H & P^ka

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 P^H = P^ka- log(HA)/(A^-)

log(HA)/(A^-) = P^ka- P^H

By putting the values log(HA)/(A^-) = P^ka- P^Hg(HA)/(A^-) = 4.25- 4

log(HA)/(A^-) = 0.25

by taking anti log (HA)/(A^-) = 10^0.25 = 1.778

(HA)/(A^-) ratio will be the protonated: deprotonated form, so answer will be 1.778

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

Biochem question Can someone explain protonateddeprotonated

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