Ah buffers... So, i need to make 1 L of a 0.06 M buffer of Histidine at pH=5.5.
Im given 1.0 M Histidine (full deprotonated), 1.0 M HCl and 1.0 M NaOH. So, right away i added 0.12 moles of HCl to bring the histidine to the right form.

So to start i used henderson hasselbach:

5.5 = 6 + log$\dfrac{A-}{HA}$
$\dfrac{A-}{HA}$=0.316

%[HA]=0.76
%[A-]=0.24

$n_{HA}$=0.76*0.06M*1L=0.0456 moles
$n_{A^-}$=0.24*0.06M*1L=0.0144 moles

SO i need to put in: $n_{HA}$=0.06 moles and $n_{NaOH}$=0.0144 moles

$V_{HA}=\dfrac{0.06moles}{1M}$
$V_{NaOH}=\dfrac{0.0144 moles}{1M}$

Question

Ah buffers...  So, i need to make 1 L of a 0.06 M buffer of Histidine at pH=5.5. 
Im given 1.0 M Histidine (full deprotonated), 1.0 M HCl and 1.0 M NaOH. So, right away i added 0.12 moles of HCl to bring the histidine to the right form.

So to start i used henderson hasselbach:

5.5 = 6 + log$\dfrac{A-}{HA}$
$\dfrac{A-}{HA}$=0.316

%[HA]=0.76
%[A-]=0.24

$n_{HA}$=0.76*0.06M*1L=0.0456 moles
$n_{A^-}$=0.24*0.06M*1L=0.0144 moles

SO i need to put in: $n_{HA}$=0.06 moles and $n_{NaOH}$=0.0144 moles

$V_{HA}=\dfrac{0.06moles}{1M}$
$V_{NaOH}=\dfrac{0.0144 moles}{1M}$

aaronq · Answer

just need someone to say this is okay, or not.