Question #133404
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 . If 10ml of 10M HCl was added to 1 litre of the above buffer (question 9) what would be the resulting pH?
What would be the resulting pH if 10ml of 10M HCl were added to 1 litre of pure water?
1
Expert's answer
2020-09-17T07:18:00-0400

pH=pKa+lg(CBACHA)pH=pK_a+lg(\frac{C_{BA}}{C_{HA}}), MOPS is a middle strength acid (pKa = 7.2)

CBACHA=10pHpKa=107.87.2=3.98\frac{C_{BA}}{C_{HA}}=10^{pH-pK_a}=10^{7.8-7.2}=3.98

On the other hand, the required concentration of the buffer is 0.4M:

CBA+CHA=0.4C_{BA}+C_{HA}=0.4

Now, we can calculate the required concentrations of MOPS and its Na-salt:

3.98CHA+CHA=0.43.98C_{HA}+C_{HA}=0.4

CHA=0.43.98+1=0.08MC_{HA}=\frac{0.4}{3.98+1}=0.08M

CBA=0.4CHA=0.32MC_{BA}=0.4-C_{HA}=0.32M

n(HA)=CHAVsol=0.08M6L=0.48moln(HA)=C_{HA}*V_{sol}=0.08M*6L=0.48mol

n(BA)=CBAVsol=0.32M6L=1.92moln(BA)=C_{BA}*V_{sol}=0.32M*6L=1.92mol

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

Now you take 2.40mol*209.3g/mol = 502g of MOPS and 1.92mol/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.



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