What is the pH of the solution created by combining 2.40 mL of the 0.10 M NaOH(aq) with with 8.00 mL of the 0.10 M HC2H3O2(aq)?
So, here's my working so far:
.00024 mol NaOH
.0008 mol HC2H3O2
.0008 - .00024 = .00056 mol HC2H3O2
.00056mol / 0.0104L = .05385 M HC2H3O2
The Ka of acetic acid in water at 25C = 1.76 x 10^-5.
So I find [H+] = 9.73 x 10^-4
and pH = 3.01, but this is not the right answer. Can anyone explain where I went wrong?
Ka is ok. (HC2H3O2) is ok.
Where is the (C2H3O2^-)? I don't see that anywhere.
NaOH + HC2H3O2 ==> NaC2H3O2 + H2O
Ka = (H^+)(C2H3O2^-)/(HC2H3O2) and solve for (H^+) = Ka*(HC2H3O2)/(C2H3O2^-)
(H^+) = ....
I get something like 4 x 10^-5 and pH about 4.4 or so. (You need to do it more accurately).
To find the correct answer, let's go through the steps again.
Step 1: Calculate the moles of NaOH and HC2H3O2 in the given volumes.
- NaOH: 2.40 mL * 0.10 M = 0.240 mmol = 0.00024 mol
- HC2H3O2: 8.00 mL * 0.10 M = 0.800 mmol = 0.00080 mol
Step 2: Determine the excess amount of HC2H3O2.
- Excess HC2H3O2 = final moles of HC2H3O2 - moles of NaOH reacted
- Excess HC2H3O2 = 0.00080 mol - 0.00024 mol = 0.00056 mol
Step 3: Calculate the new concentration of HC2H3O2.
- Total volume after mixing = 2.40 mL + 8.00 mL = 10.40 mL = 0.0104 L
- [HC2H3O2] = moles of HC2H3O2 / total volume
- [HC2H3O2] = 0.00056 mol / 0.0104 L = 0.05385 M
Step 4: Use the equilibrium expression to find [H+].
- HC2H3O2 is a weak acid, so we use the equilibrium expression: Ka = [H+][C2H3O2-] / [HC2H3O2]
- Ka = 1.76 x 10^-5, [HC2H3O2] = 0.05385 M
- Let x be the concentration of [H+]
- Ka = x^2 / (0.05385 - x)
Step 5: Solve the quadratic equation to find [H+].
- Rearrange the equation: x^2 = Ka * (0.05385 - x)
- Substitute the value of Ka: x^2 = (1.76 x 10^-5) * (0.05385 - x)
- Solve the quadratic equation and find the value of [H+]. In this case, [H+] ≈ 0.00931 M.
Step 6: Calculate the pH.
- pH = -log[H+]
- pH = -log(0.00931)
- pH ≈ 2.03
Based on these calculations, the pH of the solution should be approximately 2.03, not 3.01.