How would you prepare 05 L of a 08 M solution of citrate buf

How would you prepare 0.5 L of a 0.8 M solution of citrate buffer that has a pH of 4.50? ThepKa of citric acid is 4.74. The only solutions that you may use to accomplish this are: (1) a 1.0 Msolution of citric acid (the weak acid form), and a solution of 10.0M NaOH. You do NOT haveaccess to any other reagents. Include in your answer the volumes of each of these solutions that youwould mix together to do so. (5 marks)

pH = pKa + log ([A-]/[HA])

-0.24 = log ([A-]/[HA])

0.5457 = [A-]/[HA]

solving in M**:

0.8M = [HA] + [A-]

plug into above:

0.5457 = (0.8M – [HA])/[HA]

[HA] = 0.509

[A-] = 0.291

Thus, to make this, start with 0.5L of citric acid and add 0.291M of NaOH, which works out to:C1V1 = C2V2(10.0)(V1) = (0.5L)(0.291M)V1 = 14.55 mL NaOH

can you show me \"step by step \" how to solve for [HA]?

Solution

pH = pKa + log ([A-]/[HA]) [henderson-hasselbach equation]

pka=4.74

4.50=474+ log ([A-]/[HA])

-0.24 = log ([A-]/[HA])

0.5457 = [A-]/[HA]

Also totl buffer concentration=0.8M = [HA] + [A-]

plug into above:

0.5457 = (0.8M – [HA])/[HA]

0.5457[HA]=0.8M-[HA]

[HA]+0.5457[HA]=0.8M

1.5457[HA]=0.8M

[HA]=0.8/1.5457=0.517M

[A-] = 0.8-0.517=0.282M

Thus, in the buffer solution the concentration of citrate,[A-]=concentration of added NaOH

Citric acid(HA) +NaOH --->citrate(A-) +H2O (neutralization of citric acid was done to produce citrate )

concentration of added NaOH=10M=C1

volume of NaOH added=V1

mol of NaOH added=C1V1=10M*V1

Also ,mol of NaOH in buffer=0.282M=C2

volume of buffer=V2=0.5L

C1V1=C2V2

gives, 10M*V1=(0.282M)(0.5L)

V1=(0.282M)(0.5L)/10M=0.0141L=14.1ml