8 mL of 0.0100 M HCl are added to 22 mL of 0.0100 M acetic acid. What is the pH of the resulting solution?
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To find the pH of the resulting solution, we need to consider the equilibrium that occurs when HCl reacts with acetic acid. The balanced chemical equation for this reaction is:
HCl + CH3COOH ⇌ CH3COOH2+ + Cl-
Since HCl is a strong acid, it completely dissociates in water. Therefore, the concentration of HCl in the solution is 0.0100 M.
The acetic acid used is a weak acid, so it undergoes partial dissociation. However, for the purpose of this calculation, we can assume it dissociates completely. Hence, the concentration of acetic acid also remains 0.0100 M.
Now, let's build an ICE table (Initial, Change, Equilibrium) for this reaction.
Initial:
[HCl] = 0.0100 M
[CH3COOH] = 0.0100 M
Change:
[HCl] = -x
[CH3COOH] = -x
[CH3COOH2+] = +x
[Cl-] = +x
Equilibrium:
[HCl] = 0.0100 - x
[CH3COOH] = 0.0100 - x
[CH3COOH2+] = x
[Cl-] = x
Since x is very small compared to the initial concentration of acetic acid and HCl, we can approximate 0.0100 - x as 0.0100. This simplification makes the equation for pH calculation much easier.
In the reaction, H3O+ ions are produced by the dissociation of acetic acid:
CH3COOH ⇌ CH3COOH2+ + OH-
The dissociation constant for acetic acid is denoted as Ka. Ka = [CH3COOH2+] / [CH3COOH]. In this case, it is equal to x^2 / (0.0100 - x). We can approximate the value of Ka as x^2 / 0.0100, considering x is much smaller than 0.0100.
The equation for the ionization of water is:
H2O ⇌ H+ + OH-
Since the concentration of OH- in pure water is 1.0 x 10^-7 M, we can assume that the concentration of OH- produced by the ionization of water is negligible compared to the concentration of OH- produced by the dissociation of acetic acid.
So, we consider the concentration of OH- in the reaction as equal to the concentration of OH- produced by the dissociation of acetic acid.
Since the concentration of acetic acid is 0.0100 M, and the concentration of [OH-] is equal to x, we have:
0.0100 = x
Rearranging the equation, we find:
x = 0.0100
This tells us that the concentration of [OH-] (and [CH3COOH2+]) is 0.0100 M.
Taking the negative logarithm of the concentration of [H3O+] gives us the pH:
pH = -log[H3O+]
Since H3O+ is produced by the dissociation of HCl, which has a concentration of 0.0100 M, we can conclude that:
pH = -log(0.0100)
Using a calculator, we find:
pH ≈ 2.00
Therefore, the pH of the resulting solution is approximately 2.00.