# chemistry

Methanoic acid HCO2H(aq) also known as formic acid, is partly responsible for the characterisitic itchy rash produced by the leaves of the stinging nettle plant. Calculate the pH of 0.150 mol/L methanoic acid. The Ka for methanoic acid is 1.8 x 10^-4.

My Work:
Lets call Methanoic acid HL

HL--> H^+ plus L^-

Ka = (H+)(L-)/(HL) = 1.8 x 10^-4

HL --> H+ plus L-
I 0.150 0 0
C -x +x +x
E 0.150-x +x +x

Ka = (+x)(+x)/(0.150-x) = 1.8 x 10^-4
= (x^2) / (0.150-x) = 1.8 x 10^-4

Im having problems with the next step

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1. You appear to have more trouble solving the math problem than you do the chemistry problem.

Ka = (+x)(+x)/(0.150-x) = 1.8 x 10^-4
= (x^2) / (0.150-x) = 1.8 x 10^-4

If we try the simplifying route first, we assume 0.150-x - 0.15 and the equation becomes
x^2/0.150 = 1.8 x 10^-4
Then x^2 = 1.8 x 10^-4 x 0.150 = 2.7 x 10^-6
x = sqrt (2.7 x 10^-6) = 0.0052 and pH = -log(0.0052) = 2.28.
Now we check to see if our assumption holds. Is 0.0052 small when compared with 0.150. Yes it is for 0.158-0.0052 = 0.144 so the error is only about 3.4% error. USUALLY we can make this assumption if the error is less than 5%.

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posted by DrBob222
2. Making the assumption keeps us from solving a quadratic. However, we could solve the quadratic and the answer comes out to be 0.051 so you see that's very close to what we got with the simplifying assumption.

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posted by DrBob222
3. Thank you! I still have to read over it but for now I beleive I got the jist of it

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posted by Ross
4. In re-reading Q&A, I noticed I made a typo here. That should be 0.0051 and not 0.051.

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posted by DrBob222

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