Calculate the pH of each of the following solutions.
(a) 0.200 M propanoic acid (HC3H5O2, Ka = 1.3 10-5)
7.29 (i was correct)
(b) 0.200 M sodium propanoate (NaC3H5O2)
13.3 (i was incorrect)
(c) pure H2O
7.00 (obviously correct)
(d) 0.200 M HC3H5O2 and 0.200 M NaC3H5O2
havent tried it yet.
i dont know how to do be without ka
2007-12-01 07:11:42 · 4 answers · asked by poo 2 in Science & Mathematics ➔ Chemistry
(b) NaC3H5O2 is a strong electrolyte => Na+ + C3H5O2-
C3H5O2- + H2O <----> C3H5O2H + OH-
for this equilibrium the costant also called hydrolysis constant is Kw/Ka = 10^-14 / 1.3 x 10^-5 = 7.69 x 10^-10
7.69 x 10^-10 = x^2 / 0.200 -x
x = 1.24 x 10^-5 M = [OH-]
pOH = 4.91
pH = 9.09
(d) this is a buffer
Ka ( as you told) is 1.3 x 10^-5
pKa = 4.88
we use Hasselbalch's equation
pH = pKa + log [C3H5O2-] / [HC3H5O2]
pH = pKa + log 0.200 / 0.200 = 4.88
2007-12-01 07:30:10 · answer #1 · answered by Dr.A 7 · 2⤊ 0⤋
Ka = [H+][CH3CH2C))-] / [CH3CH2COOH]
a). When propanoic acid dissociates it forms H+ and CH3CH2COO- in equal amounts.
So the CH3CH2COO- can be substituted for another H+ in the equation.
Ka = [H+]^2 / [CH3CH2COOH]
[H+]^2 = Ka x [CH3CH2COOH]
[H+]^2 = 1.3 x 10^-5 x 0.20 = 2.6 x 10^-6
[H+] = sq rt [2.6 x 10^-6] = 1.61 x 10^-3
pH = - log(10)[H+]
pH = - log(10) [ 1.61 x 10^-3]
pH = - -2.79
pH = 2.79
2007-12-01 07:30:10 · answer #2 · answered by lenpol7 7 · 2⤊ 0⤋
I know this is 7 years ago but you can obviously tell that a. is not correct because it is propionic ACID, and you got a pH of 7.29 which makes it a very weak base, almost neutral
2015-02-09 17:17:04 · answer #3 · answered by Steven Ly 2 · 0⤊ 0⤋
is this homework? i think you should do it on your own.
2016-05-27 03:20:02 · answer #4 · answered by ? 3 · 0⤊ 4⤋