a) What is the pH of a 2.0 molar solution of acetic acid. Ka of acetic acid = 1.8 x 10¯5

(b) A buffer solution is prepared by adding 0.10 liter of 2.0 molar acetic acid solution to 0.1 liter of a 1.0 molar sodium hydroxide solution. Compute the hydrogen ion concentration of the buffer solution.

(c) Suppose that 0.10 liter of 0.50 molar hydrochloric acid is added to 0.040 liter of the buffer prepared in (b). Compute the hydrogen ion concentration of the resulting solution.

I am just wondering if my answers are right?

a) pH: 2.22
b) 1.8 x 10 ^-5
c) .214

For your first one I get 2.37
[H+][Ac-]/[HAc]=Ka since 1:1 disociation
and small qty.. the equation
x * x/(1-x) approx = X^2/1=Ka=1.8 e-5
x=.0042 =[H+]
pH=2.37 significant figures pH 2.4

your 2nd question is harder. but I agree with your results.. the NaOH neutalized 1/2 of your acetic acid.
leaving the concentration you indicated.

3rd question
you have .1 l of 0.5M/l HCL .05 moles
you have 0.04 L of 1.8e-5 [H+] a very small number
using simple dilutions Normality* volume= Normality * volume
.1 L of 0.5 N HCl= (0.1 +0.04)l * X N HCL about 0.357M/l conc of H+

For your answer to part B

the hydrogen ions (-protons) in the acid would most probably latch onto the hydroxide ions available. Thus, adding more water to the solution, and the acetic acid losing much of it's strength. Yeah, a higher pH number now. So the ionic power should increase too: somewhere between -5 and zero.

Though, it has been a long while since I was in touch with Chem......

Happy Solving!

Pin~Jinx / anarchist