# 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.

Apr 26, 2018

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 $1 \cdot {10}^{-} 14$...

${K}_{b} = \frac{{K}_{w}}{{K}_{a}}$

${K}_{b} = 7.7 \cdot {10}^{-} 10$

Use ${K}_{b}$ (to one decimal place) to solve for your change in concentration via algebra...

${K}_{b} \text{ (or "K_a")"= "Equilibrium concentration of products"/"Equilibrium concentrations of reactants}$

$7.7 \cdot {10}^{-} 10 = \frac{{x}^{2}}{0.025 - x}$

Rearrange the variables to make a quadratic equation...

$\left(1\right) {x}^{2} + \left(7.7 \cdot {10}^{-} 10\right) x - 1.925 \cdot {10}^{-} 11$
$a {x}^{2} + b x + c$

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...

$9.1 = 14 + {\log}_{10} \left(1.4 \cdot {10}^{-} 5\right)$

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

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