Determine the pH?

Question

Determine the pH of a 1.0M NaC3H5O3 if .1M solution of HC3H5O3 has a pH of 2.44

Please explain. Thanks so much!

Answer 1

Without going thru all the gory details.
(1) convert pH to [H+]. Call the result (X)
(2) Compute X^2/0.1, call result Ka
(3) Compute 10^-14 / Ka, call result Kb
(4) Compute Y for Y^2/1 = Kb. This is [OH-] for the sodium salt solution
(5) Compute pY
(6) Subtract pY from 14. That's your pH

Answer 2

This question deals with conjugate acids and bases. Here, HC3H5O3 is the acid and C3H5O3(-) (in the form of NaC3H5O3) is the conjugate base. Since we are given the pH of the solution of acid, and we need to find the pH of the base, we are going to have to find a Ka value and convert it to a Kb value. The first step is to set up an ICE chart to find the Ka value.


___HC3H5O3 <--> H(+) + C3H5O3(-)
I_____.1_________0_______0
C____-x_________+x______+x
E___.1 - x________x_______x

In this case, we can find the concentration of H(+) be using the given pH. Since this concentration is actually equal to "x" we can find all the ending concentrations. Once we have done that, we can solve for Ka.

x=10^-2.44 = .00363

Therefore:
[HC3H5O3] = .0964
[H(+)] = .00363
[C3H5O3(-)] = .00363

Ka = [C3H5O3(-)][H(+)]/[HC3H5O3]

Ka = (.00363)(.00363)/(.0964) = 1.367x10^-4

Now that we have Ka, we can convert it to Kb using this:

Kw = Ka * Kb

Kb = 1x10^-14/1.367x10^-4 = 7.315x10^-11


Now that we have Kb, we set up the follwoing equilibrium and use an ICE chart to find concentrations:

_C3H5O3(-) + H2O <-> HC3H5O3 +OH(-)
I____1________________0______0
C___-x_______________+x_____+x
E__1 - x______________x_______x

Kb=x^2/(1-x)

Since Kb is very small, ignor the "x" in the denominator.

7.315x10^-11 = (x^2)/1
x = 8.55x10^-6

Now that we have the concentration of OH(-), we can find pOH.

pOH = -log(8.55x10^-6) = 5.07

pH = 14 - pOH = 8.93

The pH of the solution would be 8.93 making it slightly basic.

Answer 3

First we must get the Ka

pH = 2.44

[H+] = 10^-2.44 = 0.00363 M

for the equilibrium

HC3H5O3 < > C3H5O3- + H+

initial concentration
0.1
at equilibrium
0.1 - 0.00363 . . . 0.00363 . . 0.00363

K = (0.00363 )^2 / 0.0964

Ka = 0.000137

Second we have an hydrolysis : NaC3H5O3 is a strong salt >> Na+ + C3H5O3-

C3H5O3- + H2O <> HC3H5O3 + OH-

The equilibrium constant is Kh = Kw / Ka = 1 x 10^-14 / 0.000137 = 7.30 x 10^-11

7.30 x 10^-11 = x^2 / 1.0 - x

x = [OH-] = 8.54 x 10^-6 M

pOH = 5.07

pH = 8.93