Question

If I dissolve 0.500 g of sodium butanate, NaC₃H₇COO, MM = 110.1 g/mol, in water to make a solution with a final volume of 250.0 mL, what will be the pH of this solution?

I started out with finding the concentration of sodium butanate using .500 g and the MM

.500g/110.2g/mol = .0182 M

Ka of butanoic acid, C₃H₇COOH, is 1.77x10^-4

I then proceeded to equate that Ka value to the [H₃O⁺][C₃H₇COOH]/[NaC₃H₇COO]

Did I set this up correctly? If not, how would you solve this problem?

Answer 1

Answer – We are given the mass of the sodium butanate = 0.500 g , molar mass = 110.1 g/mol , volume of water = 250.0 mL

First we need to calculate the moles of sodium butanate

Moles of sodium butanate = 0.500 g / 110.1 g.mol^-1

                                           = 0.00454 moles

Now we need to calculate the molarity of sodium butanate

[NaC₃H₇COO] = 0.00454 moles / 0.250 L

                         = 0.0182 M

We know, [NaC₃H₇COO] = [C₃H₇COO^-] = 0.0182 M

Ka for the butanoic acid = 1.5*10^-5

We need to put ICE chart

   C₃H₇COO^- + H₂O -----> C₃H₇COOH + OH^-

I   0.0182                             0                0

C    -x                                 +x             +x

E 0.0182-x                    +x             +x

We need to calculate the Kb from Ka

We know,

Kb = 1*10^-14 / Ka

      = 1*10^-14 / 1.5*10^-5

       = 6.66*10^-10

We know,

Kb = [C₃H₇COOH][OH^-] / [C₃H₇COO^-]

6.66*10^-10 = x*x /(0.0182-x)

We can neglect x in the 0.0182-x, since 5% rule and Kb value is too small

6.66*10^-10 * 0.0182 = x²

So, x = 3.48*10^-6

We know, x = [OH^-] = 3.48*10^-6 M

We know,

pOH = -log [OH^-]

        = - log 3.48*10^-6 M

        = 5.46

Now ,

pH + pOH = 14

so, pH = 14 – pOH

            = 14 – 5.46

             = 8.54

The pH will be for this solution is 8.54