1. You are asked to prepare a buffer of pH 6.90 for a biological assay. Using th

Question

1. You are asked to prepare a buffer of pH 6.90 for a biological assay. Using the table of pKa’s in the experimental procedure (Part 4), select an appropriate acid/base pair and clearly state the pair that you selected. Calculate the volume of the acid solution and the volume of the conjugate base solution that would be needed to prepare a buffer with a total volume of 300. mL. Assume all solutions are 1.00 M. Show all of your work. 2. Using the buffer solution prepared above in Question 1, what would be the resulting pH of the buffer after the addition of 18.0 mL of 1.20 M HNO3?

Explanation / Answer

A buffer is a solution that do not undergo change in pH(>1) when small amount of strong acid or base is added to it. When a small amount of acid is added to a buffer, then the conjugate base undergoes the neutralization reaction,

B- + H3O+BH + H2O

When a small amount of base is added to a buffer, then the conjugate base undergoes the reaction,

AH + OH-A- +H2O

And so the ratio of [base]/[acid] may change by small amount of the range 0.1 and the pH-Pka=+/-1

2)pH=pka+ log [base]/[acid] …..henderson –hasselback equation

[H2PO4-]=1.00 M=1 mol/L

[HPO42-]=1.00M=1 mol/L

6.90=7.21 +log [base]/[acid]

-0.31=log [base]/[acid]

[base]/[acid] =0.49

Total volume=300ml

[base] =0.49[acid]

Or, [HPO42-]/[H2PO4-]=0.5

Using formula,

M1*V1=M2*V2                                           M=molarity, V=volume

M1/M2=V2/V1

[HPO42-]/[H2PO4-]=1/2 =V(H2PO4-)/ V(HPO42-)

Or, V(HPO42-) = 2 *V(H2PO4-)

Or,V(base)=2 *V(acid)                                        (as 0.49=0.5(approx)=1/2)

V(base) + V(acid) =300ml

Or, 2* V(acid)+, V(acid)=300

3 V(acid)=300 ml

Or, V(acid)=100ml

And V(base)=2*100ml=200ml

pH of buffer=6.90

To calculate resultant pH we need to know the [base]/[acid]

Moles of HNO3 added=18.0 ml * 10^-3 L/ml*1.20mol/L=0.0216

Moles of

Moles of

Conjugate base is neutralized on addition of strong acid

So moles of base neutralized=0.0216

Moles of c.base remaining in the solution0.2-0.0216=0.18 moles

Conc of c.base in the solution=0.18 moles/total volume=0.18 moles/(300+18) ml=5.7*10^-4 moles/ml =5.7*10^-4 *1000=0.57 M

Conc of conjugate acid is increases by 0.0216 moles (due to conversion of c.b)

Now, moles of conjugate .acid=0.1+0.0216=0.1216 moles

Conc of conjugate acid=0.1216 moles/318 ml=3.8 *10^-4 moles/ml=3.8 *10^-4*1000 moles/L=0.38 M

pH=7.21 + log [base]/[acid]=7.21 + log 1.5=7.38

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