Question

The titration of 321 mL of 0.501 M carbonic acid (H2CO3) (Ka1 = 4.3 x 10-7,...

The titration of 321 mL of 0.501 M carbonic acid (H2CO3) (Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11) with 1.5 M KOH.

How many mL of the 1.5 M KOH are needed to raise the pH of the carbonic acid solution to a pH of 6.755?

I tried many times, but I just dont get the answers right. How do I do this?

Homework Answers

Answer #1

when we use weak acid and strong base we get buffer

pH = pka + log [conjugate base]/[acid]      is equation for pH of buffer

pka = -log Ka1 = -log ( 4.3x10^-7) = 6.3665

now we use this   6.755 = 6.3665 + log [HCO3-]/[H2CO3]

[HCO3-] = 2.446 [H2CO3]

HCO3- moles= 2.446 H2CO3 moles .............(1) ( since Molarity = Moles/vol, vol term cancells from both side)

initially H2CO3moles = M xV = 0.501 x ( 321/1000) = 0.16082

Let KOH vol be V , then KOH moles = M x V = 1.5V

KOH reacts with H2CO3 and we get HCO3- and H2O

hence H2CO3 moles after reactiong with KOH = 0.16082-1.5V

HCO3- moles formed = KOH moles reacted = 1.5V

now we use this in eq ( 1)

1.5V = 2.446 ( 0.16082-1.5V)

1.5V = 0.39337 - 3.669V

V = 0.076 L = 76 ml

hence 76 ml of 1.5 KOh is needed to rise pH to 6.755

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