Calculating charge of amino acids (am I doing it right?)

Please consider this picture of a peptide chain:


![enter image description here](https://i.stack.imgur.com/JNMa1.png)


I've done some examining on the different amino acids and concluded that the chain is as follows:


Lys-Leu-Gly-Ser-Citrullin (variant of arginine).


I would now like to calculate the charge of this peptide at neutral pH.


I'm using this table to orient myself: <http://www.sigmaaldrich.com/content/dam/sigma-aldrich/life-science/biochemicals/migrationbiochemicals1/common-amino-acids-table.jpg>


So here's my train of thought:


As we're at neutral pH the +1 and -1 of the mandatory amino and acid groups will cancel out for all amino acids.


Lysine has a pKx of barely 11 which is some higher than 7 so it is basic in our solution. It means that it will accept a proton so Lysine has a charge of +1 in our respective solution.


Leucine, Glycine and Serine have no charge on their x group so their charge is 0 and does not contribute anything to the total charge.


Citrullin, I read, is uncharged at neutral pH, opposed to Arginine which is, so it does not contribute either.


So my total answer is +1. Is this correct? I am in no way 100% sure of my method but it would be great to get a confirmation that I am doing it correctly or to be corrected on my possibly faulty method.


Thank you!




I second Lighthart's "Looks good to me"



> 
> Let's break out the [Henderson-Hasselbalch equation](http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation) to be sure.
> 
> 
> 


The Henderson-Hasselbalch equation is usually used to determine the pH of a buffer, but we can work backwards to get the ratio of an acid to its conjugate base at any pH.


$$\ce{pH}=\ce{p}K\_a +\log{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}$$
$$\log{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}=\ce{pH}-\ce{p}K\_a$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}=e^{\ce{pH}-\ce{p}K\_a}$$


Let's take the three possible cases:



> 
> $\ce{pH}>\ce{p}K\_a$
> 
> 
> 


If the pH is greater than the pKa, then 


$$\ce{pH}-\ce{p}K\_a>0$$


$$e^{\ce{pH}-\ce{p}K\_a}>1$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}>1$$
$$[\ce{A-}]>[\ce{HA}]$$


For OH and SH functional groups in proteins, when $\ce{pH}>\ce{p}K\_a$, then the group is deprotonated and has a negative charge.


For amino groups, $\ce{HA}$ is $\ce{RNH3+}$, so for amino groups at $\ce{pH}>\ce{p}K\_a$, they are not protonated and neutral.



> 
> $\ce{pH}<\ce{p}K\_a$
> 
> 
> 


If the pH is less than the pKa, then 


$$\ce{pH}-\ce{p}K\_a<0$$


$$e^{\ce{pH}-\ce{p}K\_a}<1$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}<1$$
$$[\ce{A-}]<[\ce{HA}]$$


For OH and SH functional groups in proteins, when $\ce{pH}<\ce{p}K\_a$, then the group is protonated and neutral.


For amino groups at $\ce{pH}<\ce{p}K\_a$, they are protonated and positively charged.



> 
> $\ce{pH}=\ce{p}K\_a$
> 
> 
> 


In this case you will find that:
$$[\ce{A-}]=[\ce{HA}]$$


Half of the groups are protonated and the other half are not.


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