Question

What is the pH of a 0.025 M aqueous solution of sodium propionate? NaC3H5O2(aq) + H2O(l) <---> HC3H5O2(aq) + Na+(aq) + OH-(aq) The Na+ is a spectator ion, so it can be eliminated from the equation to give: C3H5O2-(aq) + H2O(l) <---> HC3H5O2 -(aq) + OH

The Ka for propionic acid is 1.3e-5.

Answer 1

The pH should be about `9.14`

Explanation:

Hydroxide (OH[-]), is a product in this reaction, and since sodium propionate is a weak base, you have to build an ICE table (Initial-Change-Equilibrium concentration table), and find Kb from the Ka, to solve for the change in concentration. Don't include water since water is a liquid, therefore the concentration really cannot change.
Ensure you use appropriate significant figures

The I.C.E. table:

Solve for Kb by dividing the water auto-ionization constant, which is `1*10^-14`...

`K_b = (K_w) / (K_a)`

`K_b = 7.7*10^-10`

Use `K_b` (to one decimal place) to solve for your change in concentration via algebra...

`K_b " (or "K_a")"= "Equilibrium concentration of products"/"Equilibrium concentrations of reactants"`

`7.7*10^-10 = (x^2)/(0.025 - x)`

Rearrange the variables to make a quadratic equation...

`(1)x^2 + (7.7*10^-10 )x - 1.925 * 10^-11`
`ax^2+bx+c`

`x=(-b±sqrt (b^2-4ac))/(2a)`

Use the positive value of x to the appropriate significant figures to find pOH and then find pH...

`9.1=14+log_10(1.4*10^-5)`

So... the pH of your solution is 9.1!

NOTE: Since this is chemistry, use significant figures all throughout your math!