Hi guys, I need some help with this, I think I have a but b and c? Thank you 1.

Question

Hi guys, I need some help with this, I think I have a but b and c? Thank you

1. Consider the various protonation state of phosphate. a. (5 points) Draw Lewis structures to illustrate the various protonation states of phosphate. Which are the two most common species at pH 7.00. (5 points) At what percent of total phosphate are each of these two species present? You may ignore other protonation states in your calculation. (3 points) Justify the assumption that we can ignore the other protonation states in our calculation b. c.

Explanation / Answer

Phosphate ions - 1) PO43-(phosphate) ,HPO42-(hydrogen phosphate), H2PO4-(dihydrogen phophate),H3PO4(phosphorc acid)

pka1=2.12 , H3PO4 ----> H+ +H2PO4-

pka2=7.21, H2PO4- ------>H+ + HPO42-

pka3=12.68, HPO42- ----->H+ + PO43-

b) Using Henderson-hasselbach equation:

pH=pka1+log [H2PO4- ]/[H3PO4]

7.00=2.12+ log [H2PO4- ]/[H3PO4]

or log [H2PO4- ]/[H3PO4]=7.00-2.12=4.88

or, [H2PO4- ]/[H3PO4]=10^4.88(very high)

Similarly,

pH=pka2+log [HPO42-]/[H2PO4- ]

7.00=7.21 +log [HPO42-]/[H2PO4- ]

So, [HPO42-]/[H2PO4- ]=10^-0.21=0.616

and

pH=pka3+log [PO43-]/[HPO42-]

7.00=12.68 +log[PO43-]/[HPO42-]

So, [PO43-]/[HPO42-]=10^-5.68=2.089*10^-6 (negligible)

Species present :[HPO42-] and [H2PO4- ] as H3PO4 is almost totally deprotonated,

As [HPO42-]/[H2PO4- ]=0.616

Let [HPO42-]+[H2PO4- ]=X

[HPO42-]=[H2PO4- ]*0.616

So,[H2PO4- ]*0.616 +[H2PO4- ]=X

or, 1.616[[H2PO4- ]=X

or,[H2PO4- ]=X/1.616

percent [H2PO4-]={ [H2PO4-]/[H2PO4- ]+[HPO42-]} *100={(X/1.616)/X}*100=0.619*100=61.9%

So, percent [HPO42-]=100-61.9=38.1%

c) first equivalence point is reached at pH=1/2(pka1+pka2)=1/2(2.12+7.21)=4.6 So, H3PO4 is totally deprotonated to form ,H2Po4-

Second equivalence point , is reached t pH=1/2(pka2+pka3)=1/2(7.21+12.68)=9.94 ,So at pH=7.00 very less dissociation of H2Po4- to HPO4- takes place

Third equivalence point is reached at pH>>9.9, So at pH=7.00 ,negligible dissociation takes place for HPO4- to PO43- (ignored)

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