Calculating pH and pOH for 0.048 M Benzoic Acid Solution

[Coco12]

Homework Statement

what is the H, Oh , ph and poh for a 0.048 mol/l solution of benzoic acid? C6H5COOH

Homework Equations

Ph +poh=14
Ph=-log(h30+)
H30=10^-ph

The Attempt at a Solution

I thought maybe u need to separate benzoic acid into its ions and then use the molar ratio to find the concentration of H? but what does that break down into?

 

[Borek]

C₆H₅COOH <-> C₆H₅COO^- + H⁺

Note: it is a weak acid, so you need to use its dissociation constant to calculate pH.

 

[Coco12]

Ok we did not learn that yet but I will look it up in my chemistry book. Thanks

 

[epenguin]

Ok we did not learn that yet but I will look it up in my chemistry book. Thanks

Better still, if you haven't done it already, try and work it out without looking it up in your chemistry book. You know something about equilibrium equations and mass conservation. If you succeed you will gain confidence and understanding, if you don't succeed the explanation will mean more and be better learned when you do look it up and see maybe what key point you'd missed. If you've looked it up already, apply this principle in future problems in chemistry or any other science as far as possible. It will even make things easier for you in the long run.

 

[DeeNos]

To calculate the pH and pOH for a 0.048 M solution of benzoic acid (C6H5COOH), we first need to understand the dissociation of this acid in water. Benzoic acid is a weak acid, meaning it does not fully dissociate in water. It will partially dissociate into its ions: C6H5COOH ⇌ C6H5COO- + H+.

Using the equilibrium constant (Ka) for benzoic acid, we can calculate the concentration of H+ ions in the solution. Ka for benzoic acid is 6.5x10^-5, so we can set up an equilibrium expression:

Ka = [C6H5COO-][H+] / [C6H5COOH]

Since we know the concentration of benzoic acid (0.048 M), we can plug that in and solve for [H+]:

6.5x10^-5 = (x)(x) / (0.048-x)

Where x is the concentration of H+ ions. Solving for x, we get 0.0063 M. This is the concentration of H+ ions in the solution.

To find the pH, we use the equation pH = -log[H+]. Plugging in our concentration of H+, we get a pH of 2.2.

Since we know that pH + pOH = 14, we can also calculate the pOH by subtracting the pH from 14. So pOH = 14 - 2.2 = 11.8.

To find the concentration of OH- ions, we use the equation pOH = -log[OH-]. Plugging in our pOH value, we get a concentration of OH- ions of 0.000063 M.

In summary, for a 0.048 M solution of benzoic acid, the pH is 2.2, the pOH is 11.8, the concentration of H+ ions is 0.0063 M, and the concentration of OH- ions is 0.000063 M.