# 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

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The Ka for propionic acid is 1.3e-5.

The Ka for propionic acid is 1.3e-5.

##### 1 Answer

The pH should be about

#### 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#...**

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

**Rearrange the variables to make a quadratic equation...**

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

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

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