Dissociation of glycine

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In summary, the conversation discusses the problem of determining the amount of 5M KOH needed to be added to 1L of 0.1M glycine solution at a pH of 9 to reach a pH of 10. The answer given was 10mL, which the person did not understand. The solution involves using the Henderson-Hasselbach equation to calculate the moles of R-NH2 at pH 9, then using the pH equation to find the increase in moles at pH 10. The resulting value of 13mL is not the correct answer, but may be due to rounding. The conversation also touches on the confusion of glycine nomenclature and the use of the ratio of base
  • #1
apchemstudent
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Here's the problem how much of 5M KOH should be added to 1L of 0.1M glycine(pKa=9.6) at a pH of 9 until the pH reaches exactly 10? the answer in the back says 10mL.

R-NH3(+) <-> R-NH2 + H(+)

Here's the way I solved it. I figured out the number of moles of R-NH2 using the henderson-hasselbach equation, and it is .251 when the pH was 9.

And then i figured out the increase in moles of R-NH2 when the pH is at 10:

pH = pKa + log(NH2+x / NH3(+) - x)
10 = 9.6 + log((moles of NH2+x)/(moles of NH3(+) - x))

x = .064

then for every mole of R-NH2 dissociated a mole of KOH must be used.

so concentration * volume = .064
5M * V = .064
V ~ 13 mL

This is not the correct answer, although it might seem close. How did they end up with 10 mL? I tried rounding it to 0.06 but that only gave me 12 mL. Am i being too precise here, and the book just rounded down to 10 mL due to significant figures? Please help. Thanks
 
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  • #2
You're solution seems fine, if your not convinced try working backwards, you'll see that the the 13mL corresponds to the pH of 10.
 
  • #3
Thanks for your help
 
  • #4
ahh I get it, the .1M glycine doesn't represent the acid, it represents glycine, the molecule, as a whole. The nomenclature's all **** up. Well, there's biochemistry for ya, nomenclature in genetics is also a travesty. Unfortunately I don't believe they have the name for the different protonated forms of glycine.

the .251 is the ratio of the base to the acid, to find the concentration of the base

[tex](.251/1.251).1~=~initial~base~conc.[/tex]
[tex](.1/1.251)=~initial~acid~conc.[/tex]

[tex]K=[.02006+x]10^{-10}/(.07994-x)=2.51x10^{-10}[/tex]

[tex]x=.05145=v5/(v+1)[/tex]

[tex]v=~10mL[/tex]

If anyone needs any clearing up, just say so, I admit the solution's pretty messy.
 

1. What is glycine?

Glycine is a non-essential amino acid, meaning that it can be produced by the body and is not required in the diet. It is the simplest amino acid, with a hydrogen atom as its side chain.

2. What is dissociation of glycine?

Dissociation of glycine refers to the breakdown of glycine into its component parts, which are an amino group (NH2) and a carboxyl group (COOH). This process occurs when glycine is dissolved in water, and the amino and carboxyl groups become ionized, creating a zwitterion.

3. What is the pH at which glycine is fully dissociated?

The pH at which glycine is fully dissociated (meaning that all molecules have been broken down into their component parts) is known as the isoelectric point, and for glycine it is at a pH of 5.97.

4. How does temperature affect the dissociation of glycine?

The dissociation of glycine is an endothermic process, meaning that it absorbs heat. This means that as temperature increases, the rate of dissociation also increases. However, at very high temperatures, the amino and carboxyl groups can be damaged, leading to a decrease in the rate of dissociation.

5. What is the significance of studying the dissociation of glycine?

Understanding the dissociation of glycine is important in many fields, including biochemistry, medicine, and food science. It can help us understand how amino acids function in the body, how they can be used as building blocks for proteins, and how they contribute to the taste and nutritional value of foods.

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