Calculate [OH^-]in a solution obtained by adding 0.0100 mol solid NaOH to 1.00 L of 15.0 MHH3.

Name: Problem 179, Chapter 13: Chemistry An Atoms First Approach
Uploaded: 2020-05-11T15:26:13-08:00
Duration: 3 min 37 s
Channel: Anatole Borisov
Description: Calculate [OH^-]in a solution obtained by adding 0.0100 mol solid NaOH to 1.00 L of 15.0 MHH3.

Calculate [OH^-]in a solution obtained by adding 0.0100 mol...

Key Concepts

Strong Base Dissociation

A strong base, such as NaOH, dissociates completely in water, providing an immediate increase in the hydroxide ion concentration. This complete ionization contrasts with weak bases and is critical for determining the overall ion concentration when mixed with other basic or acidic species.

Weak Base Equilibrium

Weak bases, like ammonia (NH3), do not fully ionize in water. Instead, they establish an equilibrium between the undissociated base and the ions produced (NH4? and OH?). This equilibrium is governed by the base’s dissociation constant (Kb), which quantitatively describes the extent of ionization.

Common Ion Effect

The common ion effect occurs when an ion that is a part of the equilibrium is added to the solution from an external source. In this case, the addition of OH? from NaOH shifts the equilibrium of the weak base (NH3) dissociation, thereby suppressing its further ionization. Understanding this effect is essential when calculating ion concentrations in mixed solutions.

Equilibrium Constant (Kb)

The equilibrium constant for a weak base represents the ratio of the concentrations of the products of ionization (NH4? and OH?) to that of the un-ionized base (NH3). This constant is used in setting up the equilibrium expression, which can lead to a quadratic equation when additional sources of common ions are present in the solution.

Transcript

00:01 It's just a really interesting question that asks about how the equilibrium of nh3 and water with oh minus and nh4 plus.

00:13 So basically, nh3 acting as a base, how it's affected by the addition of 0 .01 moles of oh minus, i think in the form of naoh.

00:26 So that's basically just saying oh minus because it dissociates 100 % in a 1 -liter solution.

00:31 So right off the bat, i'm just going to say, because it's in a one -liter solution, and this is added in solid form, it's not going to change the volume.

00:39 And so we don't need to worry about it as moles or something.

00:44 We can just think of it as 0 .01 molar.

00:47 And that's going to be easier because in ice tables, in mass action expressions, we always work in concentration.

00:53 So what we're really saying is for the equilibrium, which we're going to do an ice table for, ignore water because it's not part of the mass action expression.

01:02 The initial concentration of oh h minus is going to be 0 .01 molar for nh4 plus it's going to be 0 and for nh3 it's going to be 15 molar because that's what we were given as the initial concentration of n h3 and so it's just like a normal one except instead of having this uh zez 0 over here it's 0 .01 so now because we couldn't possibly go to the left because there's no nh4 plus we know the reaction is going to go to the right so minus x over here as the change plus x plus x and then 12 minus x 0 .01 plus x and so if we plug this into a mass action expression for so ka is equal to on the top we'll have point 0 .01 plus x times x so and on the bottom we'll have 15 minus x.

02:06 Now, commonly, we will ignore this value of x when it's in a sum like this to make the problem simpler.

02:13 However, in this case, it'll actually turn out that you can't do that.

02:19 And the way you can recognize this just like visually is that we have .01 isn't that big...

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