Consider sodium acrylate, NaC3 H3 O2 . Ka for acrylic acid (its conjugate acid) is 5.5 ×10^-5. (

Consider sodium acrylate, NaC3 H3 O2 . Ka for acrylic acid...

Consider sodium acrylate, $\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2} . K_{\mathrm{a}}$ for acrylic acid (its conjugate acid) is $5.5 \times 10^{-5}$. (a) Write a balanced net ionic equation for the reaction that makes aqueous solutions of sodium acrylate basic. (b) Calculate $K_{\mathrm{b}}$ for the reaction in (a). (c) Find the $\mathrm{pH}$ of a solution prepared by dissolving $1.61 \mathrm{~g} \mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}$ in enough water to make $835 \mathrm{~mL}$ of solution.

Consider sodium acrylate, $\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2} . K_{\mathrm{a}}$ for acrylic acid (its conjugate acid) is $5.5 \times 10^{-5}$.
(a) Write a balanced net ionic equation for the reaction that makes aqueous solutions of sodium acrylate basic.
(b) Calculate $K_{\mathrm{b}}$ for the reaction in (a).
(c) Find the $\mathrm{pH}$ of a solution prepared by dissolving $1.61 \mathrm{~g} \mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}$ in enough water to make $835 \mathrm{~mL}$
of solution.

Key Concepts

pH Calculation in Weak Base Solutions

Determining the pH of a solution that contains a weak base involves calculating the concentration of hydroxide ions produced by the equilibrium reaction of the weak base with water. This requires converting a given mass of salt into molarity, applying the equilibrium constant (Kb) for the reaction, and then solving for the hydroxide ion concentration. Finally, the pOH is calculated from [OH–] and converted to pH, which is an important procedure in the study of acid–base equilibria.

Salt Hydrolysis

This concept explains that certain salts, when dissolved in water, undergo a reaction called hydrolysis, where the anion (or cation) reacts with water to produce a weak acid (or weak base) and hydroxide (or hydronium) ions. In cases where the salt comes from a weak acid, its conjugate base will react with water to generate OH–, making the solution basic. This is the underlying principle behind why aqueous solutions of such salts are basic and forms the basis for writing a net ionic equation for the reaction.

Acid–Base Conjugate Pairs

Acid–base conjugate pairs are species that transform into each other by the gain or loss of a proton. In the context of salt hydrolysis, the anion from a weak acid (the conjugate base) interacts with water to form the weak acid and hydroxide ions. Understanding this relationship is crucial to determine which species is active in altering the pH of a solution and how equilibrium is established in the reaction.

Equilibrium Constant Relationships

This concept refers to the interconnection between the acid dissociation constant (Ka) and the base dissociation constant (Kb) for conjugate acid–base pairs, connected via the water ion product constant (Kw = 1.0×10?¹? at room temperature). Specifically, Kb can be determined by the relationship Kb = Kw/Ka. Recognizing this relationship is essential for calculating the base strength of the conjugate base resulting from the salt and for solving equilibrium problems in aqueous solutions.

Transcript

00:01 So this question asked us to take a look at some characteristics of acrylic acid, which is h .c .3 .h .302, i believe.

00:16 Yes, that's correct.

00:17 And we are given first the ka of this, which is equal to 5 .6 times 10 to negative 5th.

00:27 And we are asked to figure out first what the ph of this solution would be if it has a 0 .1 molar concentration.

00:37 So this is a really straightforward problem that we've seen many times.

00:41 We'll write the ka here.

00:44 Ka equals 5 .6 times 10 to negative fifth is equal to the concentration of hydronium plus i'll just write a minus for the conjugate base.

00:57 Of acrylic acid over h .a., which if you do an ice table for, and if you need to do the ice table, pause the video and do it, but the result will be that we'll get a value of x squared over 0 .1 minus x, where x is equal to the concentration of h3 plus and a minus.

01:17 And since x is gonna be much smaller than point one, we'll ignore it in this case.

01:21 And so just doing some simple plugging into a calculator, we're gonna get that x is equal to 2 .36 times 10 to negative third, which i'll copy that over, that 2 .36 times 10 to negative third, value is equal to the concentration of h3o plus, which then the ph we can figure out as the negative log of this value, which is the definition of how the relationship of ph and hydrogenium concentration is, and we'll get that that equals 2 .62.

01:55 So that's our answer for part one.

01:58 The next part asks us, what is the percent dissociation of that 0 .1 molar acrylic acid? so let's go back to here.

02:08 What we're looking at is, actually, i'll write out the equation generically as h .a.

02:15 For acrylic acid reacts with water to form hydrogenium plus the conjugate acid.

02:22 And so the percent dissociation is basically asking what portion of this one molar h .a.

02:28 Goes and becomes h3o plus and a minus.

02:35 And so in order to figure that out, we can take the concentration of h3o plus that we have at equilibrium and divide that by this initial concentration, 0 .1.

02:45 And so we actually already calculated this concentration of h3o plus.

02:49 That was our value of x right here...

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