given .100 M of HC3H5O2 and Ka is 1.3x10^-5 what is PH?

Question

hsa to do with buffer solutions category. propanoic acid is name of compund.

Answer 1

HC3H5O2 <> C3H5O2- + H+
initial concentration
0.100....... .. .... ..... 0.............0

at equilibrium
0.100-x............ .. ....x... .... .....x

1.3 10^-5 = (x)(x) / 0.100-x

x = 0.00114 M

pH= - log 0.00114 = 2.94

Answer 2

hi,

This is the fmethod for usually working out the pH of a weak acid. Propanoic Acid is a weak acid because it partially dissociates in aqueous solution to form:

HX ⇌ H+ and X-

So the equation to calculate H+ from this is:

[HX] = Propanoic Acid
[H+] = Protons
[X-] = Rest of the Acid
[ ] = These Square brackets represent the concentration of what is inside them.

This is the Dissociation Constant: Ka = [H+][X-] / [HX]

So to find [H+]: Rearrange to form:

[H+][A-] = Ka x [HX]

And you must assume: [H+] is equal to[A-] at equilibrium, in that case the equation becomes :

[H+]^2 = Ka x [HX]

And therefore to find [H+] = √ (Ka x [HX] )

(√ is Square Root of)

In your question:

[H+] is equal to = √ (( 1.3 x10^-5) x (0.100))

Which gives: 1.14 x 10 ^-3 (Mol dm -3)

So to find pH

pH= -Log [H+] (When you use this, make sure you dont remove the number from your calculator and use the full answer you get.)

Therfore pH is : 2.94

But you had mentioned something about Buffer solutions, becuase i only Ka and the concentration of Propanoic acid i used a formula where i knew two of the three required variables, and so i would have done it this way, if i am wrong, please forgive me.

Need more help with these equations relating to acids and organic chem, ask me...i will try, my best.

Cheers

Mussa

Answer 3

Take the negative logarithm of the root of (Ka x 0.100).