Answer to Question #133403 in Inorganic Chemistry for Dennielle

Question

Six litres of 0.4M 3-(N-morpholino) propane sulfonic acid (MOPS) buffer of pH 7.8 are required. Available in the laboratory are 10M HCl and 10M NaOH and the zwitterionic form of MOPS (M.Wt = 210.3). The pKa for the MOPS zwitter ion is 7.2. Describe, showing the relevant calculations, the procedure for preparing the buffer

Answer 1

pH = pKa ​+ log (C_BA / C_HA)

7.8 = 7.2 + log (C_BA / C_HA)

log (C_BA / C_HA) = 0.6

So:

C_BA / C_HA = 3.98

and C_BA + C_HA = 0.4

Solving this system of equations, we have:

C_HA ​= 0.08M

C_BA ​= 0.4 − C_HA ​= 0.32M

$n(HA)=C_{HA}*V_{sol}=0.08M*6L=0.48mol$

$n(BA)=C_{BA}*V_{sol}=0.32M*6L=1.92mol$

$n(HA)+n(BA)=0.48mol+1.92mol=2.40mol$

Now we take 2.40 mol * 209.3 g/mol = 502 g of MOPS 

and 1.92 mol/10M = 0.192L of 10M solution of NaOH (react in a 1 to 1 molar relation):

C₇H₁₅NO₄S + NaOH = C₇H₁₄NO₄SNa + H₂O

After the reaction is complete, you dissolve this mixture up to 6L in a special flask with distilled water.