# is this right?--ionization - (Aug/11/2007 )

hi there

can someone please see if i have the right answer...hv a test tomorrow and just wanna make sure im doing it right..

its assumed that 2mL of 1M NaOH added to 6mL of a diluted solution of tyrosine fully ionises the side chain Tyr-O^-, and that 1M HCl gives a fully protonated side chain Tyr-OH. the absorbance at 296nm of the fully ionised form=0.712 and the absorbance for the protonated form was -0.038. after addition of 2mL of 180mL of NaOH the absorbance was 0.593. the resulting % ionization is?

and so i did...0.593/0.712*100=83.3%

now to me this seems right but its a multi choice question and my answer isn't there..

the options present are

a-84.1%

b-79.1%

c-100%

d-80%

e-50%

maybe a is the right answer? sumone please help

Thanks in advance

b_06er

can someone please see if i have the right answer...hv a test tomorrow and just wanna make sure im doing it right..

its assumed that 2mL of 1M NaOH added to 6mL of a diluted solution of tyrosine fully ionises the side chain Tyr-O^-, and that 1M HCl gives a fully protonated side chain Tyr-OH. the absorbance at 296nm of the fully ionised form=0.712 and the absorbance for the protonated form was -0.038. after addition of 2mL of 180mL of NaOH the absorbance was 0.593. the resulting % ionization is?

and so i did...0.593/0.712*100=83.3%

now to me this seems right but its a multi choice question and my answer isn't there..

the options present are

a-84.1%

b-79.1%

c-100%

d-80%

e-50%

maybe a is the right answer? sumone please help

Thanks in advance

b_06er

Hi!

Your mistake is in zero determination. 0.712 - is 100% ionisation point , and -0.038 - 0% ionisation point , so all the meaning scale is 0.712+0.038 = 0.75

So You should do so (0.593/0.75)*100 = 79.1% Answer b