What is the pH of the resulting solution if 30.00 mL of 0.100M acetic acid is added to 10.00mL of 0.100 M NaOH? For acetic acid, Ka=0.000018.

I know that NaOH is a strong base and acetic acid is a weak acid. What is the equation: C6H5COOH+NaOH<--> C6H5COOH
Once I get this, I think I'll be able to use Ka from there. Thanks.

The equation is

CH3COOH + NaOH ==> CH3COONa + H2O
(What you wrote--C6H5COOH-- is benzoic acid).
I would convert M and L to mols and use the Henderson-Hasselbalch equation. Post your work if you get stuck.

To determine the pH of the resulting solution, we can start by writing the balanced chemical equation for the reaction between acetic acid (C6H5COOH) and sodium hydroxide (NaOH). The reaction goes as follows:

C6H5COOH + NaOH → C6H5COONa + H2O

Given that acetic acid is a weak acid and sodium hydroxide is a strong base, we assume that the reaction goes to completion and the sodium acetate (C6H5COONa) dissociates completely in water. Therefore, the equation can be simplified to:

C6H5COOH + OH- → C6H5COO- + H2O

Next, we can determine the initial concentrations of acetic acid (C6H5COOH) and hydroxide ion (OH-) in the solution.

For acetic acid:
Initial concentration = 0.100 M x (30.00 mL / (30.00 mL + 10.00 mL)) = 0.075 M

For hydroxide ion:
Initial concentration = 0.100 M x (10.00 mL / (30.00 mL + 10.00 mL)) = 0.025 M

Now, we can use the equilibrium constant expression for the dissociation of acetic acid to calculate the concentration of acetate ion (C6H5COO-) and hydronium ion (H3O+). The dissociation of acetic acid is given by:

C6H5COOH + H2O ⇌ C6H5COO- + H3O+

Ka = [C6H5COO-][H3O+]/[C6H5COOH]

Rearranging the equation, we have:

[H3O+] = (Ka [C6H5COOH])/[C6H5COO-]

Substituting the given values, we have:

[H3O+] = (0.000018 M x 0.075 M)/0.025 M

Simplifying, we find:

[H3O+] = 0.000054 M

Finally, we can calculate the pH using the equation:

pH = -log[H3O+]

Substituting the value we just found, we have:

pH = -log(0.000054)

Calculating this value, we get the pH of the resulting solution.

Note: In this calculation, we assumed that the volumes are additive and that there is no change in volume upon mixing the solutions.