11 mL of 0.0100 M HCl are added to 25 mL of 0.0100 M acetic acid. What is the pH of the resulting solution?

Calculate the (H^+) contributed by the acetic acid remembering that the HCl will suppress the ionization of the acetic acid. Then add the H^+ contributed by the HCl.. Acetic acid is HAc.

(HAc) = 0.0100 x 25/36 = approx 0.007 M
(HCl) = 0.0100 x 11/36 = 0.003 M
....................HAc ==> H^+ + Ac^-
I.................0.007.....0.003.......0
C.................-x.............x...........x
E............0.007- x....0.003+x....x

Then Ka = (H^+)(Ac^-)/(HAc) = (0.003+x)(x)/(0.007-x)
Solve for x, then total H^+ = 0.003+x

Post your work if you get stuck.

To find the pH of the resulting solution, we need to calculate the concentration of hydronium ions (H3O+) in the solution.

First, let's determine the reaction between HCl and acetic acid (CH3COOH):

HCl + CH3COOH -> CH3COOH2+ + Cl-

Since acetic acid is a weak acid, it doesn't completely dissociate in water. Therefore, we need to consider the equilibrium of acetic acid:

CH3COOH ⇌ CH3COO- + H3O+

Given that the initial concentration of HCl is 0.0100 M and the volume is 11 mL, the moles of HCl can be calculated using the formula:

moles = concentration x volume

moles of HCl = 0.0100 M x (11 mL / 1000 mL) = 0.00011 mol

Now, let's calculate the moles of acetic acid using the same calculation:

moles of acetic acid = 0.0100 M x (25 mL / 1000 mL) = 0.00025 mol

Since the moles are in a 1:1 ratio for HCl and acetic acid, it means that all of the HCl will react with the acetic acid. Therefore, the moles of acetic acid remaining can be calculated by subtracting the moles of HCl:

moles of acetic acid remaining = moles of acetic acid - moles of HCl = 0.00025 mol - 0.00011 mol = 0.00014 mol

Now, we can calculate the concentration of hydronium ions (H3O+) in the resulting solution. Since the remaining acetic acid is in equilibrium with the hydronium ions, we can use the equilibrium expression:

Ka = [CH3COO-] / [H3O+]
where Ka is the acid dissociation constant for acetic acid.

Given that the Ka for acetic acid is 1.8x10^-5, we can substitute the values into the equation:

1.8x10^-5 = 0.00014 mol / (0.0100 L - 0.011 L)

Simplifying the equation:

1.8x10^-5 = 0.00014 mol / 0.099 L

Rearranging the equation to solve for [H3O+]:

[H3O+] = (0.00014 mol / 0.099 L) x 1.8x10^-5

[H3O+] = 2.5714x10^-7 M

Finally, to find the pH, we need to take the negative logarithm of the hydronium ion concentration:

pH = -log10(2.5714x10^-7)

Therefore, the pH of the resulting solution is approximately 6.59.