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
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%.
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.
Thank you! I still have to read over it but for now I beleive I got the jist of it
In re-reading Q&A, I noticed I made a typo here. That should be 0.0051 and not 0.051.
To solve for the pH of the solution, we need to find the concentration of H+ ions. From the balanced equation, we can see that the concentration of H+ ions is equal to the concentration of x (the change in concentration of H+ ions).
Thus, we have the equation:
Ka = (x^2)/(0.150 - x) = 1.8 x 10^-4
Now we need to solve for x, the change in concentration of H+ ions. But since Ka is small, we can assume that x is much smaller than 0.150, and therefore we can approximate 0.150 - x as approximately 0.150.
So we have:
(1.8 x 10^-4) = (x^2)/(0.150)
Now we can rearrange the equation to solve for x:
x^2 = (1.8 x 10^-4)(0.150)
x^2 = 2.7 x 10^-5
Taking the square root of both sides:
x = √(2.7 x 10^-5)
x ≈ 0.0052
Since we assumed that x is small, we can approximate the concentration of H+ ions as x ≈ 0.0052 mol/L.
Now, to find the pH, we use the equation:
pH = -log[H+]
pH = -log(0.0052) ≈ 2.28
Therefore, the pH of the solution with a concentration of 0.150 mol/L methanoic acid is approximately 2.28.