What is the pH of a 0.050 M solution of formic acid, p Ka=3.75 ?

Name: Problem 51, Chapter 2: Organic Chemistry
Uploaded: 2020-05-08T04:41:47-08:00
Duration: 8 min 3 s
Channel: Bobby Barnes
Description: What is the pH of a 0.050 M solution of formic acid, p Ka=3.75 ?

What is the pH of a 0.050 M solution of formic acid, p...

Key Concepts

pH is a measure of the acidity or basicity of a solution defined as the negative logarithm of the hydrogen ion concentration. It provides a convenient way to express the concentration of H+ ions in a logarithmic scale, making it easier to handle the wide range of possible values in chemical systems.

Weak Acid Equilibrium

Weak acid equilibrium refers to the state in which a weak acid only partially dissociates in solution, establishing a dynamic balance between the undissociated acid and its ions. This equilibrium is essential to determine the concentrations of the acid and its conjugate base using the acid dissociation constant.

Acid Dissociation Constant (Ka) and pKa

The acid dissociation constant, Ka, quantifies the extent to which an acid can donate protons in solution, serving as a measure of acid strength. The pKa is the negative logarithm of Ka and provides a more manageable number to compare the strengths of different acids; lower pKa values indicate stronger acids.

Equilibrium Calculations

Equilibrium calculations involve setting up and solving the equilibrium expression for the dissociation of weak acids or bases. These calculations often require approximations based on the degree of dissociation, especially when the acid does not ionize significantly, allowing for simpler mathematical treatment of the system.

Transcript

00:01 We want to find the ph of a 0 .05 molar solution of formic acid, given that its pca is 3 .75.

00:09 Now, the equation that we're going to want to end up using eventually is we know that ka should be equal to the concentration of hydrogen times the concentration of, well, actually, let's write this reaction out first, and then we can write this.

00:32 This a little bit better.

00:34 So if we were to react formic acid, formic acid, with water, and so remember this is going to be liquid, when these react, we'll get the adion for this.

00:56 So formic ion, and then we would have hydronium.

01:08 And this is the same thing as just h -plus.

01:11 So on the other side here, we're going to have our formic ion.

01:20 And then in the denominator, well, the only thing we're going to have, because remember we don't count h2s or h2o, since it's a pure substance, we would just have our formic acid.

01:35 So we need to figure out what are the concentrations of these.

01:38 And we also need to figure out, well, what is our k -a? so let's go ahead and figure out the k -a first.

01:47 Now, in one of the previous questions, we actually showed that first, this is just a negative log of k .a.

02:00 So if we were to undo all of our algebra and everything, we should end up getting that ka is equal to 10 raised to the negative 3 .75.

02:10 So we have our ka.

02:12 Now, to get our concentrations, well, we're going to need to set up an ice table.

02:18 So let's see.

02:20 So our initial for formic acid is 0 .050.

02:25 So i put an eye here.

02:27 There's some amount of water.

02:30 We'll just say it's an excess.

02:32 So this doesn't really matter all that much.

02:35 But then for formic acid, we're going to have zero and zero, because we're assuming we're starting with nothing.

02:42 Now the change in this, oh, actually, maybe i shouldn't use an x.

02:47 I'll just say it's not really important.

02:49 What we're doing.

02:50 So just cross all that up.

02:52 Now, the change is just going to be some amount x.

02:57 So we don't know how much is going to form.

03:00 So if we have x form, well, for every one mole formic acid we have, we should form one formic ion and one hydronium.

03:10 So this is going to be 0 .05 minus x...

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