pH of Weak Bases - Video Tutorials & Practice Problems
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1
concept
ICE Charts of Weak Bases
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51s
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Now recall that weak bases represent weak electrolytes that only partially dissociate into aqueous ions. You're going to say that they require the use of an ice chart. So ice stands for initial change equilibrium and we use this ice chart to calculate equilibrium amounts. Now here, the units of an ice chart will be in molarity, which remember is moles per liter. And we use capital M to represent molarity. And since we're dealing with weak bases, we'd use our base association constant, which is KB. So just keep in mind when it comes to weak basis because they're weak species, we have to utilize an ice chart in order to tell what their exact equilibrium amounts will be in a given process.
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example
pH of Weak Bases Example
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6m
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Here, it says to calculate the hydroxide ion concentration of a 0.55 molar potassium fluoride solution at 25 °C. Here we're told the acid association constant of HF Hydrofluoric acid is 3.5 times 10 to the negative four. All right. So step one is we're gonna set up an ice chart for the weak base that hasn't reacting with water. Now, we know that this is a weak face because this is an ionic compound. So we break it up into its ions. So we have K positive and F minus based on our understanding of ionic salts, we know that potassium is a main group. Metal, main group metals need to be plus three or higher in terms of charge in order to be acidic. Since it doesn't meet that minimum requirement, it's neutral F minus. On the other hand, adding H plus two, it creates HF which is a weak acid, which means that this F minus is basic. But since it comes from a weak acid, it is stronger, but it's still relatively weak, right? So F minus here is our weak base. Now for ionic bases, which this one is we ignore the neutral metal C. Here, we ignore the potassium. Now use the Bronson Ori definition to predict the products formed, make sure that KB is used in the presence of the weak base. So right now, we are only given the K A of its acid form, but we're gonna need to utilize its KB later on to find our final answer. So remember we're going to set up our equation. So F minus here will react with water, which is a liquid. Since it is a base, water is going to act as the acid water would therefore give an H plus over to F minus. This would create HF aqueous plus oh minus aqueous as products step two use the initial row. So remember for this, I stands for initial change equilibrium using the initial ro place the amount given for the weak base. So the weak base we're told it was 0.55 molar place a zero for any substance not given an initial amount. In a nice chart. We ignore solids and liquids. So the water is ignored. We're not told anything about our products initially. So they're both zero step three, we lose reactants to make products. So using the change row place A minus X for the reactants and A plus X for the products. So minus X, since we're losing it and plus X for the products, since we're making them, we bring down everything. So 0.55 minus X plus X plus X. Now using the equilibrium rope set up the equilibrium constant expression with A B again because it's a weak base and solve for X. We checked as if a shorter can be utilized to avoid the quadratic formula. All right. So the shortcut that we're gonna utilize is called the 500 approximation method. Basically, we're gonna take the ratio of initial concentration of our weak base and divide it by our KB value. If we get a number greater than 500 then we can ignore the minus X within our equilibrium expression. Right? So here we're gonna say KW equals K A times KB, we need to isolate KB. So KB equals KW divided by K A. So this would be 1.0 times 10 to the negative 14 divided by RK A which is 3.5 times 10 to the negative four. Doing that gives us a KB value of 2.857 times 10 to the negative 11. All right. So our initial concentration is 0.55 molar. We found our KB as 2.857 times 10 to the negative 11. When we punch that in, we get 1.925 times 10 to the 10. So we get a number much greater than 500 which means we'll be able to ignore the minus X within our equilibrium expression. Remember your equilibrium expression for this weak basis KB equals products over reactants. So we plug in the products over reactant water is liquids, we ignore it. So bringing down this expression and inserting values for what we have a we said is 2.857 times 10 to the negative 11. Both the products are X at equilibrium. So multiplied together as X squared and this will be 0.55 minus X. Well, remember we just did the ratio of initial concentration divided by KB and we saw a number much greater than 500 which means we can ignore this minus X and avoid the quadratic formula that we have here. So then all it becomes is cross multiplying my KB with 0.55. So I'm gonna do that up here. So when I do that, I'm gonna X squared equals 1.57135 times 10 to the negative 11. I don't want X squared. I just want X, we'll take the square to both sides. X equals 3.96 times 10 to the negative six molar. Now, this is our answer because we're asked to find the hydroxide ion concentration and at equilibrium that's equal to X. So we just found out what my X value is, which is equivalent to my hydroxide ion concentration. So my final answer would be 3.96 times 10 to the negative six molar.
3
example
pH of Weak Bases Example
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8m
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Here, we can say that the Ph or POH of a weak base can be calculated once the equilibrium concentration of the hydroxide ion is found. Now, this is determined by using the equilibrium row of an ice chart. In this example question, it says, what is the ph of a 0.12 molar ethyl amine solution? We're told that the KB value of ethyl lamine is 5.6 times 10 to the negative four. All right. So we're gonna start out by following the steps 1 to 3 and setting up, setting up an ice chart. So here we have our Elaine the Elaine being a weak base because its KB is less than one would react with liquid water. Since it's the base water will act as the acid donating an H plus to it. This H plus would go towards the nitrogen giving us C two H five NH three positive. This is the ethyl ammonium ion war losing an H plus would become oh minus ion. We have our equation and this is an ice chart which is initial change equilibrium. For our initial role. We're gonna place the initial amount given for athel amine which is 0.12 molar liquids and solids are ignored within a nice charge. So water will be ignored. Our products initially are zero since they haven't formed yet. And we're losing reactants minus X to make products plus X plus X, we bring down everything. So we have 0.12 minus X plus X plus X. This leads us to step four where using the equilibrium row, set up the equilibrium constant expression and sulfur X here are equilibrium constant expression. Since this is a base, we have KB equals products over reactants. So it equals ethyl ammonium ion times hydroxide ion over alaine. Here we check to see if a shortcut can be utilized to avoid the quadratic formula. The shortcut that we utilize is called the 500 approximation method. In it, we take the ratio of our initial concentration of our weak base. In this case, two are KB value. And if that ratio happens to be greater than 500 we can ignore the minus X in terms of our equilibrium constant expression. So here we have our initial concentration of 0.12 molar divided by the KB value of the weak base. When we do that, it only gives me 214.3 as a value. Since the number is not greater than 500 we cannot ignore the minus X, which means we'll have to utilize the quadratic formula which you can remember remember is negative B plus or minus square root of B squared minus four AC over two A. So come up here, our expression is 5.6 times 10 to the minus four for the weak base. And that be X squared over 0.12 minus X cross multiply those. So when we do that, we're gonna get 5.6 times 10 to the minus four. And in parenthesis 0.12 minus X and that equals X squared, distribute, distribute. So that give me 5.6 times, oops 5.6 times negative four times 10 to negative four times 0.12 gives me 6.72 times 10 to the negative five minus 5.6 times 10 to the minus four X and this equals X squared. Now my two X variables, the X squared one has the larger power. So it's our lead term. So you would add 5.6 times 10 to the minus four X to both sides. You subtract 6.72 times 10 to the negative five on both sides. When we do that, we have our new equation. So X squared plus 5.6 times 10 to the negative four, X minus 6.72 times 10 to the minus five. Here, this equation would equal A B and C setting up our quadratic formula. We have negative 5.6 times 10 to the minus four plus or minus the square root of 5.6 times 10 to the minus four squared minus four times. There's a one here, one times negative. Don't forget the negative sign negative 6.72 times 10 to the minus five divided by two times one. When we solve for this, realize that we're gonna figure everything in here and then take the square root of that answer. When we do that, we get negative 5.6 times 10 to the minus four plus or minus 0.0164 divided by two. So here there are two outcomes for X because it's plus or minus. So X here could equal negative 5.6 times 10 to the minus four plus 0.0164 divided by two or X could equal negative 5.6 times 10 to the minus four minus 0.0164 divided by two. This gives us two answers for X initial. So in the first top one, we get 0.00792 moller. And in this bottom one, we get negative 0.00848 moller. But which one is the correct one? The correct X is the X that where wherever you place it in terms of the equilibrium role, you're gonna get a positive answer. That means this negative one will not work because this negative one, if I placed it here or here would give me a negative concentration at equilibrium which is not possible, right? So here the X that we've just found, takes us to step five, the X variable that we just found would give us our hydroxide ion concentration. Since we know hydroxide ion concentration, we could find POH So we take that number that we just found. We plug it in here 0.00792. That would give me 2.10 as my po. And then remember that Ph equals negative log of H plus and that would give you my ph or you could also remember that Ph plus POH equals 14. So Ph equals 14 minus po So here would be 14 minus 2.10. And if it it's 2.10 that would come out to 11 0.90. So we'd have 11.90 as our Ph for this solution. OK? So just remember when we find X, look on the equilibrium row that X will give you either H 30 plus or in this case oh minus, you just gotta check since it gives us oh minus, we find POH and from there subtract it from 14 to find Ph.
4
concept
Calculating Percent Ionization
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1m
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When it comes to calculating the percent ionization or percent association, it's important to remember that weak bases also represent weak electrolytes that only partially ionize or dissociate into aqueous ions. Because of this fact, weak bases ionize less than 100% strong bases. On the other hand, they represent strong electrolytes and therefore ionize completely. So they would ionize 100% if we want to determine the percent ionization or percent association for a weak base. We utilize the following formula. Here we're gonna say percent ionization and percent dissociation equals the concentration of hydroxide ion at equilibrium divided by the initial concentration of my weak base times 100. Utilizing this formula will be able to determine the percent ionization of any given weak base.
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example
pH of Weak Bases Example
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6m
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Calculate the percent ionization when 73.2 g of sodium hypo iodide are dissolved with 500 MLS of solution. We're told that the K value of Hypo iota acid, which is Hio is 2.3 times 10 to the negative 11. All right. So we're going to say that Hypo iota acid is a weak acid since its K A is less than one NAIO represents its conjugate base. It has one less hydrogen because it's the conjugate base. It represents the weak base. So we're going to have to use the steps from 1 to 3 to set up our ice chart for this weak base. Remember, for ionic bases, we ignore the neutral metal cion. In this case, the sodium ion, this leaves us with the hypo iodide ion alone. Since it's a base, it will react with water. Water here will be in its liquid phase. It is the base. So water acts as the acid, meaning that water will donate an H plus to it. This creates Hio plus oh minus s products we're dealing with the week base. So this is our ice chart set up. So we have initial change equilibrium. Now, remember the units of an ice chart with a weak base have to be in molarity. So we're gonna take the 73.2 g of sodium iodide. We're gonna convert that into molds. So we're gonna stay here for every one mole of sodium iodide. We have a mass of 1 65.89 g of sodium hypo iodide. That's the mass of the sodium, the iodine and oxygen together. So here grants cancel out. And we're gonna have here for every one mole, we have a sodium iodide, we have one mole of iodide ion which equals 0.441 moles. We need a molarity for the ice chart. So molarity here will be moles over liters. So be 0.441 moles divided by 0.500 L which comes out to 0.882 molar. Now, with the ice chart, we ignore solids and liquids. So the water would be ignored. We don't know anything about the products initially. So there's zero, we lose reactants in order to make product, it bring down everything. Now here using the equilibrium role, set up the equilibrium constant expression and solve for X. Here we'll check if a shortcut can be utilized and if so we can avoid the quadratic formula. Since this is a weak base, we're gonna have to utilize KB baby would equal products over reactants. However, we're not given KB initial, we're giving K A. So we have to remember that K A times KD equals KW. So we're gonna say KB here equals KW divided by K A. So here this would be 1.0 times 10 to the minus 14 divided by RK A which we're told is 2.3 times 10 to the negative 11. Doing this gives SKB value, our KB value here would equal 4.3 times 10 to the negative four. Now that we know the KB value, we can do the 500 approximation method when the ratio of the initial concentration divided by in this case, KB, if the ratio happens to be greater than 500 then we'll be able to ignore the minus X within our equilibrium expression. So our initial concentration is 0.882 molar. And again, our KB is 4.3 times 10 to the negative four. When we punch that in that gives us 2051.2. So we got a number greater than 500. So we can ignore the minus X within our equilibrium expression. So here's our equilibrium expression when we plug in the values. And again, because the ratio is greater than 500 I can ignore this minus X. So coming up here, we can say this is X squared divided by 0.882. By ignoring the minus sex, we can avoid the quadratic formula. We're going to cross multiply these two. So when we do that we're gonna X squared equals 3.7 times 10 to the minus four. And then take the square root of both sides. X will give me approximately 0.01947. When we find X here, X gives me oh minus concentration. So that takes us to step five, we would use the X variable, which we just found to help us calculate the percent ionization of the association. So remember here, we're gonna say percent ionization or percent association for weak base equals the init the uh concentration of hydroxide ion at equilibrium which we find is 0.01947 molar divided by the initial concentration of my base, which is 0.882 molar times 100. So when we do that, we get 2.21%. So this would represent the person ionization of my hypo iodide.
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Problem
Problem
Determine the pH of a solution made by dissolving 6.1 g of sodium cyanide, NaCN, in enough water to make a 500.0 mL of solution. (MW of NaCN = 49.01 g/mol). The Ka value of HCN is 4.9 × 10−10.
A
2.648
B
13.389
C
5.294
D
11.352
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Problem
Problem
An unknown weak base has an initial concentration of 0.750 M with a pH of 8.03. Calculate its equilibrium base constant.
A
9.35 × 10−9
B
1.53 × 10−12
C
6.54 × 10−3
D
1.07 × 10−6
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