Calculate the pH of 0.34 M citric acid?

Question

Citric acid (H3C6H5O7) is a triprotic acid with Ka1 = 8.4 10-4, Ka2 = 1.8 10-5, and Ka3 = 4.0 10-6. Calculate the pH of 0.34 M citric acid

I get an answer of 1.79637 but including sig figs, i round it to 1.8 but that's wrong. what am i doing wrong?

Answer 1

Ka1 = [H+] [H2C6H5O7] / [H3C6H5O7]

8.4e-4 =  [x] [x] / [0.34 -x]

8.4e-4 [0.34 -x]=  x2

2.856e-4  - 8.4e-4x = x2

x2 &  8.4e-4x - 2.856e-4 = 0

http://www.1728.com/quadratc.htm

x = [H+] = 0.016483 molar = 0.0165 molar

pH = 1.783 which rounds off to your 1.8  (2 sigfigs)

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now the second K doesn't need the quadratic, but  how much would it change the answer?

Ka2  = [H+] [HC6H5O7] / [HC6H5O7]

1.8e-5 =  [0.0165 + x] [x] / [0.0165 -x]

which simplifies to :

1.8e-5 =  [0.0165] [x] / [0.0165]

X = 1.8e-5

when you add that [H+] to the K1's of 0.165 = 0.1652  

which won't be enough to change the answer

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which comes back as ... what's going wrong?

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could it be the K1 is wrong ..

my book has 7.4e-4

the CRC handbook has 7.2e-3

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so what is the pH if k = 7.4e-4

7.4e-4 [0.34 -x]=  x2

2.516e-4  - 7.4e-4x = x2

x2 &  7.4e-4x - 2.516e-4 = 0

http://www.1728.com/quadratc.htm

x = [H+] = 0.0155 molar

pH = 1.81 which rounds off to your 1.8  (2 sigfigs)

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ok how about

if K = 7.2e-4

7.2e-4 [0.34 -x]=  x2

2.45e-4  - 7.2e-4x = x2

x2 &  7.2e-4x - 2.45e-4 = 0

http://www.1728.com/quadratc.htm

x = [H+] = 0.0153 molar

pH = 1.82 which rounds off to your 1.8  (2 sigfigs)

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well....

i'm voting that you're right & some thing else isn't