Question

Hey there! My professor wrote this vague problem on the board and I can't figure out how to solve it so --- I apologize in advance for any lack of clarity.

50.0 mL of 0.03M of Cysteine is titrated by 0.06M of NaOH

pKa 2= 8.37

pKa 1 = 1.92

What is the pH of the solution.....

1) Before any titrant is added

2) 12.5 mL of NaOH

3) 25.0 mL of NaOH

4) 37.5 mL of NaOH

5) 50.0 mL of NaOH

6) 60.0 mL of NaOH

I really want to understand this type of problem for exams in the future so any explanation to go along with your work would be super appreciated. Thank you!!

Answer 1

cysteine proton dissociation depends on the acidity of the proton: that which is most acidic (lower pKa) will dissociate first. Consequently, the H⁺ on the ?-COOH group (pKa1) will dissociate before that on the ?-NH₃ group (pKa2).

i have shown here general dissociatio of animo acid

now we move t question,

(i) cystine is neutral in nature so pH before addition will be 7, but since its given that cystine has two pKa this means that the nitrogen is protonated and hence the cystine is acidic in nature(YOU ARE USING CYSTINE HYDROCHLORIDE) .pH is 1.52 (-log(0.03)) because moles of cystine = moles of hcl in the solution, since cystine itself is neutral the pH will be in response to the HCl there

(ii) so as we now know first the COOH will dissociate as the pKa1 is low,

milliequvalents of base added =12.5 x 0.06, =0.75.

R-COOH-----> RCOO^- + H⁺ (where R is the everything ecxept COOH of cysteine)

concentraion of acid = 0.03 x 50 =1.5 meq

concentraio of salt = concentratio of base = 0.75 meq

so now amount of COOH left undissociated =1.5-0.75, = 0.75

pH = pKa + log (salt/acid)

pH = 1.92 + log (0.75/0.75)

1.92

(b)

when 25 ml is added the concentraion of salt = 1.5 meq

and that of acid is negilible

so the pH shoots up at this point ie..this is equivalence point whucxh has pH = (pKa1 + pKa2)/2, =5.145

(c) since now whole of COOH has been dissociated , now comes the chence of the protonated amine to disscociate

as

we know that out of 37.5,25 ml is utilized to nuetralize the COOH, now 12.5ml is added, whichis 0.06 x 12.5,=0.75meq

amount of acid left = 1.5-.75, =0.75

now pH = pKa + log (salt/acid)

pH = 8.37 + log (0.75/0.75)

pH = pKa =8.37

(d)

now 25 ml is added to neutralise the protonated amine

now the pH become pka1 +pKa2 = 10.29 (not so sure about this step...sorry)

e)

now excess of base added = 10x 0.06 =0.6meq

but since its dissolved in now 110 ml of solution

hence its molarity becomes, 0.6/110 ,=5.45 x 10^-3

pOH = -log [OH^-]

or pOH = 2.26

or pH = 14-2.26

11.74