A beaker is labeled: 1.2 M sodium benzoate. What is the pH of this solution? Ka for benzoic acid (the conjug

Ka for benzoic acid (the conjugate acid of benzoate) = 6.4 x 10-5

Can you please show me how to set up the problem and how to complete it step-by-step? I would really like to understand this stuff!! Thank you.

Sodium benzoate is soluble, so it dissociates completely:

NaB --> Na+ +B-

(formula for benzoate is long, so it'll be B or B-)

So what you REALLY have in the jug is 1.2M Na+ and 1.2M B-.

Now find the Kb for benzoate:

Kw = Ka*Kb
1*10^-14 = 6.4*10^-5 (x)
Kb = 1.56*10^-10

Use that in the equilibrium reaction:

H2O + B- <--> OH- + HB

[B-] = 1.2M
[OH-] = 0
[HB] = 0

In the reaction, a certain amount of B- will react with water to produce the same amount of OH- and HB, so the concentrations change.

[B-] = 1.2 - x
[OH-] = 0 + x
[HB] = 0 + x

Plug that into the Kb expression

1.56*10^-10 = x*x/(1.2-x)
1.56*10^-10 = x^2/(1.2-x)

after some messy algebra (or the SOLVE function on the TI-89)...

x = 1.37*10^-5 = [OH-]

pOH = -log[OH-]
pOH = -log(1.37*10^-5)
pOH = 4.86

pH = 14 - pOH
pH = 14 - 4.86
pH = 9.14

Some solved sample problems of the type you are asking about are located here:

http://www.chemteam.info/AcidBase/Calc-s…