What mass of NaOH(s) must be added to 1.0 L of 0.050 M NH3 to ensure that the percent ionization

step 1 All right, guys. We're going to be doing problem 184 of Chapter 14. So they want to know what ma

What mass of NaOH(s) must be added to 1.0 L of 0.050 M NH3...

What mass of $\mathrm{NaOH}(s)$ must be added to $1.0 \mathrm{L}$ of $0.050 \mathrm{M}$ $\mathrm{NH}_{3}$ to ensure that the percent ionization of $\mathrm{NH}_{3}$ is no greater than $0.0010 \% ?$ Assume no volume change on addition of NaOH.

Key Concepts

Percent Ionization

Percent ionization measures the fraction of a weak acid or base that becomes ionized when dissolved in solution. It is calculated as the concentration of ionized species divided by the initial concentration of the weak acid or base, multiplied by 100, and is critical in understanding the strength of weak electrolytes.

Weak Base Equilibrium

Weak bases, such as ammonia, do not fully ionize in solution but instead establish an equilibrium between the un-ionized base and its ionized forms (its conjugate acid and hydroxide ions). This equilibrium is characterized by a relatively small base ionization constant (Kb), which quantifies the extent of ionization.

Common Ion Effect

The common ion effect describes how the addition of an ion already present in the equilibrium mixture shifts the equilibrium position. In the case of a weak base, adding a substance that supplies additional hydroxide ions decreases the ionization of the base by Le Chatelier’s principle, thus lowering the percent ionization.

Stoichiometric Mass Calculation

Stoichiometric mass calculation involves using molar relationships to convert between moles of a substance and its mass. In equilibrium problems, this often means determining the amount of a substance needed (in moles) to achieve a desired shift in equilibrium, and then converting that amount to grams using the substance's molar mass.

Transcript

0:00 All right, guys.

00:01 We're going to be doing problem 184 of chapter 14.

00:03 So they want to know what mass of sodium hydroxide must be added to one liter of 0 .05 molar nh3 to ensure the percent ionization of nh3 is no greater than 0 .0010 percent.

00:18 So no volume change, change on addition of nahoh.

00:22 So when nh3 reacts of water, it produces nh4 plus oh minus.

00:27 So if we have a 0 .02 ,000, 010 ionization rate, then that's assuming that that's small enough that we can assume that none of the hydroxide in our solution would be produced from this reaction.

00:51 It's just so small that it's just really insignificant.

00:55 The vast majority of hydroxides are going to be coming from, is there going to already existence solution or it's going to come from our existing solution or it's going to come from our n -a -o -h.

01:07 So we write the equilibrium equation k -b is equal to hydroxide times ammonium over ammonia.

01:14 And since we, there's a very little ionization, less than 5 % ionization, we can assume us nh3 here.

01:23 That's going to be the same as our initial nh3.

01:29 It's going to be the same as the initial concentration and equilibrium concentration.

01:35 It's going to be difference is going to be so small it's not going to matter so now don't forget that ammonia concentration over ammonia concentration that's going to create times 100 that's going to equal 0 .0014 .4%.

02:09 And so we divide this by 100 and they'll give us one, 1 times 10 to negative 5th...

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