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  • - Our goal is to find the pH of different salt solutions,

  • and we'll start with this solution of sodium acetate.

  • So in solution, we're gonna have sodium ions, Na+,

  • and acetate anions, CH3COO-,

  • and the sodium cations aren't going to react with water,

  • but the acetate anions will.

  • So the acetate anion is the conjugate base to acetic acid.

  • So, the acetate anion is going to react with water,

  • and it's gonna function as a base:

  • it's going to take a proton from water.

  • So if you add an H+ to CH3COO-, you get CH3COOH.

  • And if you take a proton away from water,

  • if you take an H+ away from H2O,

  • you get OH-, or the hydroxide ion.

  • Alright, so let's go ahead and write

  • our initial concentrations here.

  • So our goal is to calculate the pH of our solution.

  • And we're starting with .25 molar

  • concentration of sodium acetate.

  • And so that's the same concentration of our

  • acetate anion, here, so we're gonna write:

  • 0.25 molar, for the initial concentration

  • of the acetate anion.

  • And if we pretend like nothing has reacted,

  • we should have a zero concentration

  • for both of our products, right?

  • So a zero concentration for our two products.

  • Next, we think about the change.

  • So CH3COO-, the acetate anion, when it reacts,

  • is gonna turn into: CH3COOH, or acetic acid.

  • So whatever concentration we lose for the acetate anion,

  • we gain for acetic acid.

  • So if we make the concentration

  • of the acetate anion, X, that reacts...

  • Alright, so, X reacts.

  • If X concentration reacts, we're going to lose X,

  • and we're going to gain X over here, alright?

  • And it's the same thing for hydroxide.

  • We'll be gaining X, a concentration for the hydroxide.

  • So, at equilibrium, the concentration of acetate

  • would be .25 - X, so we're assuming everything

  • comes through equilibrium, here.

  • The concentration of acetic acid would be X.

  • The concentration of hydroxide would also be X.

  • Alright, next we write our equilibrium expression,

  • and since this is acetate functioning as a base,

  • we would write "Kb" here; so we write: Kb is equal to

  • concentration of our products over

  • concentration of our reactives.

  • So we have the concentration of CH3COOH times

  • the concentration of hydroxide,

  • so times the concentration of OH-

  • this is all: over the concentration of our reactants,

  • and once again, we ignore water.

  • So we have only the concentration

  • of acetate to worry about here.

  • So we put in the concentration of acetate.

  • Alright, so...

  • Let's think about the concentration

  • of acetic acid at equilibrium.

  • Alright, so at equilibrium, the concentration is X.

  • And so I go over here and put "X",

  • and then for hydroxide, it's the same thing, right?

  • The concentration of hydroxide at equilibrium is also X,

  • and so I put "X" in over here.

  • And then, for the concentration of acetate

  • at equilibrium, concentration of acetate

  • is zero point two five minus X.

  • So, 0.25 - X.

  • Next, we need to think about the Kb value for this reaction,

  • and you will probably not be able to find this in any table,

  • but you can find the Ka for acetic acid.

  • So, acetic acid and acetate is a conjugate acid-base pair,

  • and the Ka value for acetic acid

  • is easily found in most text books,

  • and the Ka value is equal to 1.8 x 10-5.

  • And our goal is to find the Kb.

  • What is the Kb for the conjugate base?

  • Now, we know that for a conjugate acid-base pair,

  • Ka times Kb is equal to Kw,

  • the ionization constant for water.

  • So we can go ahead and plug in:

  • 1.8 x 10-5 x Kb is equal to,

  • we know this value is 1.0 x 10-14.

  • So we just need to solve for Kb.

  • So we can get out the calculator here

  • and take 1.0 x 1014, and then we divide by:

  • 1.8 x 105; so we get: 5.6 x 10-10.

  • So let's get some more space down here and let's write that.

  • So Kb is equal to 5.6 x 10-10.

  • And this is equal to X squared,

  • equal to X2 over .25 - X.

  • Next, to make the math easier,

  • we're going to assume that the concentration, X,

  • is much, much smaller than .25, and if that's the case,

  • if this is an extremely small number,

  • we can just pretend like it's pretty close to zero,

  • and so .25 - X is pretty much the same thing as 0.25.

  • So let's make that assumption,

  • once again, to make our life easier.

  • So this is 5.6 x 10-10 = X2 over 0.25

  • So now we need to solve for X.

  • So we have: 5.6 x 10-10 and we're going to multiply that,

  • we're going to multiply that, by .25...

  • And so we get 1.4 x 10-10.

  • So we now need to take the square root of that number

  • and we get: X is equal to, this gives us:

  • X is equal to 1.2 times 10 to the negative five.

  • So let's go ahead and write that here.

  • So: X = 1.2 x 10-5

  • Alright, what did X represent?

  • We have all these calculations written here,

  • we might have forgotten what X represents.

  • X represents the concentration of hydroxide ions.

  • So X is equal to the concentration of hydroxide ions.

  • So let's go ahead and write that down.

  • X is equal to the; this is molarity,

  • this is the concentration of hydroxide ions,

  • and if we know that, we can eventually get to the pH.

  • That was our original question:

  • to calculate the pH of our solution.

  • So, we could find the pOH from here.

  • I know the pOH is equal to the negative log

  • of the hydroxide ion concentration.

  • So I could take the negative log of what we just got,

  • so, the negative log of 1.2 x 10-5,

  • and that will give me the pOH.

  • So let's go ahead and do that.

  • So: -log(1.2 x 10-5)

  • is going to give me a pOH of 4.92

  • So I go ahead and write:

  • pOH = 4.92

  • And finally, to find the pH, I need to use one more thing,

  • 'cause the pH + the pOH is equal to 14.

  • So I can plug in the pOH into here,

  • and then subtract that from 14.

  • So the pH is equal to 14 - 4.92

  • and that comes out to 9.08

  • So the pH = 9.08

  • So we're dealing with a basic solution for our salts.

  • Let's do another one.

  • Our goal is to calculate the pH

  • of a .050 molar solution of ammonium chloride.

  • So, for ammonium chloride, we have NH4+ and Cl-

  • The chloride anions aren't going to react appreciably

  • with water, but the ammonium ions will.

  • So let's our reaction here.

  • So NH4+ is going to function as an acid.

  • It's going to donate a proton to H2O.

  • So if H2O accepts a proton, that turns

  • into hydronium ions, so H3O+

  • And if NH4+ loses a proton, we're left with NH3

  • So let's start with our initial concentrations.

  • Well, we're trying to find the pH of our solution, and we're

  • starting with .050 molar solution of ammonium chloride.

  • So that's the same concentration of ammonium ions, right?

  • So this is .050 molar.

  • And if we pretend like this reaction hasn't happened yet,

  • our concentration of our products is zero.

  • Next, we think about the change,

  • and since NH4+ turns into NH3, whatever we

  • lose for NH4+ is what we gain for NH3.

  • So if we lose a certain concentration of X for ammonium,

  • if we lose a certain concentration of X for NH4+

  • we gain the same concentration, X, for NH3

  • And therefore, we've also gained the same

  • concentration for hydronium as well.

  • So at equilibrium, our concentration of ammonium would be:

  • .050 - X; for the hydronium ion, it would be X;

  • and for ammonia, NH3, it would be X as well.

  • So we're talking about ammonium acting as an acid here,

  • and so we're gonna write an equilibrium expression.

  • We're gonna write Ka.

  • So Ka is equal to: concentration of products over reactants,

  • so this would be the concentration of:

  • H3O+ times the concentration of NH3

  • all over, the concentration of NH4+

  • 'cause we're leaving water out,

  • so, all over the concentration of NH4+

  • Alright, the concentration of hydronium ions at equilibrium

  • is X, so we put an "X" in here.

  • Same thing for the concentration of NH3

  • That would be X, so we put an "X" into here.

  • This is all over, the concentration of ammonium,

  • which is .050 - X.

  • So over here, we put 0.050 - X.

  • Next, we need to think about the Ka value.

  • So finding the Ka for this reaction is usually not

  • something you would find in a table in a text book.

  • But we know that we're talking about an acid-base,

  • a conjugate acid-base pair, here.

  • So, NH4+ and NH3 are a conjugate acid-base pair.

  • We're trying to find the Ka for NH4+

  • And again, that's not usually found in most text books,

  • but the Kb value for NH3, is.

  • It's: 1.8 times 10 to the negative five.

  • So for a conjugate acid-base pair,

  • Ka times Kb is equal to Kw.

  • We're trying to find Ka.

  • We know Kb is 1.8 x 10-5

  • This is equal to: 1.0 times 10 to the negative 14.

  • So we can once again find Ka on our calculator.

  • So, 1.0 x 10-14...

  • We divide that by 1.8 x 10-5

  • And so, the Ka value is: 5.6 x 10-10

  • So if we get some room down here,

  • we say: Ka = 5.6 x 10-10

  • This is equal to: so it'd be X squared over here...

  • And once again, we're going to assume that X

  • is much, much smaller than .050

  • So we don't have to worry about X right here,

  • but it's an extremely small number,

  • .050 - X is pretty much the same as .050

  • So we plug this in and we have: .050, here.

  • So we need to solve for X.

  • We get out the calculator, and we're going to take

  • 5.6 x 10-10, and we're going to multiply by .05

  • and then we're gonna take the square root

  • of that to get us what X is.

  • So X is equal to 5.3 times 10 to the negative six.

  • So: X = 5.3 x 10-6

  • X represents the concentration of hydronium ions,

  • so this is a concentration, right?

  • This is the concentration of hydronium ions,

  • so to find the pH, all we have to do

  • is take the negative log of that.

  • So, the pH is equal to the negative log

  • of the concentration of hydronium ions.

  • So we can just plug that into here:

  • 5.3 x 10-6, and we can solve; and let's take the

  • - log(5.3 x 10-6)

  • And so we get: 5.28, if we round up, here.

  • So let me get a little more room...

  • So we're rounding up to 5.28 for our final pH.

  • So pH = 5.28

  • So we got an acetic solution, which is what we would expect

  • if we think about the salts that we

  • were originally given for this problem.

- Our goal is to find the pH of different salt solutions,

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B2 中高級 美國腔

鹽溶液的pH值 (pH of salt solutions)

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    Wayne Lin 發佈於 2021 年 01 月 14 日
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