glycine and lysine, have the following values of the relevant acid dissociation constants (pKa)
glycine, pKa=2.35
lysine,pKa=10.79
For an aqueous solution of glycine alone, calculate the value of pH at which the ratio of the concentration of neutral glycine zwitterions to the concentration of protonated cation is 102.
This tutorial may help.
http://www.chem.uiuc.edu/CLCtutorials/104/AminoAcidTitrations/SeeIt.html
I would use the Henderson-Hasselbalch equation and substitute into the log term either 1/100 or 100/1 depending upon which way you want to write it.
Here is the titration curve for glycine.
https://www.google.com/search?q=titration+amino+acids&client=firefox-a&hs=ha9&rls=org.mozilla:en-US:official&channel=sb&tbm=isch&tbo=u&source=univ&sa=X&ei=jjr9U9jcKdH9yQTljIH4Dw&ved=0CCYQsAQ&biw=1024&bih=609#facrc=_&imgdii=_&imgrc=HsTtE2YEggYW1M%253A%3BEMsM7v-IV9G1fM%3Bhttp%253A%252F%252Fwww.chem.fsu.edu%252Fchemlab%252Fbch4053l%252Fcharacter%252Ftitration%252Findex_clip_image006.jpg%3Bhttp%253A%252F%252Fwww.chem.fsu.edu%252Fchemlab%252Fbch4053l%252Fcharacter%252Ftitration%252F%3B369%3B275
thank you I'm looking into this albeit totally lost atm
9.78+log10(1/102)
=7.77
sounds good?
Nevermind, I did it, I should be paying you for all this help ;)
I had to take other pk
2.35+log10(102)
=4.35
4.35 is correct
To find the value of pH at which the ratio of the concentration of neutral glycine zwitterions to the concentration of protonated cation is 10^2, we need to consider the acid dissociation equilibrium of glycine.
The acid dissociation reactions for glycine can be represented as follows:
Glycine (H2NCH2COOH) ⇌ H+ (H3N+CH2COOH-) + A- (H2NCH2COO-)
The equilibrium constant expression for this reaction can be written as:
Ka = ([H3N+CH2COOH-] * [H2NCH2COO-]) / [H2NCH2COOH]
The ratio of the concentration of neutral glycine zwitterions to the concentration of protonated cation can be expressed as:
[H2NCH2COO-] / [H3N+CH2COOH-] = 10^2
Let's assume the concentration of glycine (H2NCH2COOH) is C, which is equal to the concentration of the protonated cation ([H3N+CH2COOH-]). Therefore, the concentration of the neutral glycine zwitterions ([H2NCH2COO-]) would be 10^2 * C.
Now, we can substitute these values into the equilibrium constant expression:
Ka = ([H3N+CH2COOH-] * [H2NCH2COO-]) / [H2NCH2COOH]
= (C * (10^2 * C)) / C^2
= 100
The equilibrium constant (Ka) for glycine is related to the acid dissociation constant (pKa) by the equation:
pKa = -log10(Ka)
Substituting the given pKa value of glycine (pKa = 2.35) into the equation above, we can calculate the value of Ka:
Ka = 10^(-pKa)
= 10^(-2.35)
Now, we can equate this value to 100 and solve for the concentration of glycine (C):
10^(-2.35) = 100
-2.35 = log10(100)
-2.35 = 2
Therefore, the concentration of glycine (C) is equal to 2 M.
To find the pH at which the ratio is 10^2, we need to determine the concentration of H+ ions. For a solution of glycine at 2 M, the concentration of H+ ions can be determined using the equation:
[H+] = C / (1 + 10^(pKa - pH))
Substituting the values, we get:
[H+] = 2 / (1 + 10^(2.35 - pH))
Now, we can equate [H+] to the concentration of H+ ions that corresponds to pH 2:
10^(2.35 - 2) = 2 / (1 + 10^(2.35 - pH))
10^0.35 = 2 / (1 + 10^(2.35 - pH))
Simplifying further, we get:
1.307 = 2 / (1 + 10^(2.35 - pH))
Rearranging the equation, we get:
1 + 10^(2.35 - pH) = 2 / 1.307
Simplifying, we get:
10^(2.35 - pH) = 1.529
Taking the logarithm of both sides, we get:
2.35 - pH = log10(1.529)
Rearranging the equation, we get:
pH = 2.35 - log10(1.529)
Calculating this value, we find:
pH ≈ 1.76
Therefore, the pH at which the ratio of the concentration of neutral glycine zwitterions to the concentration of protonated cation is 10^2 is approximately pH 1.76.