pH of Weak Acids - Video Tutorials & Practice Problems
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concept
ICE Charts of Weak Acids
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44s
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Now, we're calling that weak acids represent weak electrolytes that only part and associate into ions. You're going to say they require the use of an ice chart. Now, ice here stands for initial change equilibrium. So they require the use of an ice chart in order to calculate equilibrium amounts. We're also going to say that the units of an ice chart for weak acid will be in molarity. And because it's a weak acid, we'll use K A which is our acid association constant. So keep this in mind when we're asked to find the equilibrium amounts for any potential weak acid.
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example
pH of Weak Acids Example
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6m
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Here in this example question, it says calculate the hydro num ion concentration for 0.30 molar of hydro cyanic acid. The acid association constant K A for hydro cyanic acid is 4.9 times 10 to the negative 10. All right. So what we need to realize first is that hydro cyanic acid is a weak acid. And we know this because it's K A is less than one. Remember if your acid has a K less than one, it is a weak acid. And they're asking us to calculate the hydro ion concentration which is H plus concentration or H +30 plus concentration to do that. To calculate this equilibrium amount. We follow the, the, the following setup. So step zero, we use the following steps when asked to determine the equilibrium concentration of any compound in your equation. Remember the brackets there mean concentration here. Step one, we set up an ice truck for the weak acid that has it reacting with water. So here goes our hydro cyanic acid, we're gonna react it with water here. What will be in its liquid face? Now use the Bronson Laurie definition to predict the products formed make sure that K A is used in the presence of weak acid. We don't have to worry about that because they gave us K A in the very beginning using the bros and Lauri definition. Remember that your hydro cyanic acid is an acid and that means your water will act as a base. Remember the acid donates in H plus. So when it does that, we're gonna get CN minus being created since it's an ion, it's aqueous water except in H plus become H +30 plus also aqueous. So here goes our equation for the dissociation of our weak acid. Now, here we're, we've set up a nice chart. So we're gonna say we have initial change equilibrium. Now using the initial roll place the amount given for the weak acid place as zero for any substance not given an initial amount. All right. So we're told within the question that we're dealing with 0.30 molar of hydro cyanic acid. Remember in a nice chart, we ignore solids and liquids. Water is a liquid. So water will be ignored. We're not told anything about the initial amounts of our products. So they're both initially zero step three, we lose reactants to make products. So using the change row place A minus X for the reactants and third, we're losing them and A plus X for the products since we're making them. So this would be minus X plus X plus X. Now using the equilibrium Ro set up the equilibrium constant expression with K A. So remember we're using K A and solve for X. Now here for the equilibrim row, we just bring down everything. So this is gonna be 0.30 minus X plus X plus X. Now here we're gonna check to see if a shortcut can be utilized to avoid the quadratic formula here to do that. We're gonna use what's called the 500 approximation method in it, we're gonna use the ratio of initial concentration to the K that we're using. In this question. In this case, it would be K A. And if that ratio is greater than 500 then we could ignore the minus X within our equilibrium expression. So here our initial concentration given to us for hydro cyanic acid within the question was 0.30 molar. The K A of this weak acid is 4.9 times 10 to the negative 10. When we punch that in it gives me a number of 6.12 times 10 to the eight. So this number is way larger than 500 which means I can ignore the minus X within my equilibrium expression. Remember your equilibrium expression, we come back up here. Remember that is K A equals products over reactants. So it equals CN minus times H 30 plus divided by HCN. Again, water is a liquid. We ignore solids and liquids. That's why it's not there. If we plug in the numbers that we want for it. So K A is 4.9 times 10 to negative 10. Both of my products are X. So multiplied together, that would be X squared. And then at equilibrium, my HCN is 0.30 minus X. By using the 500 approximation method, we saw that our ratio was greater than 500. So I could ignore this minus X. And because of that, I could ignore the quadratic formula. But just for reference, remember, the quadratic formula here would be negative B plus or minus B squared minus four ac over two A. So this would be our quadratic formula. Luckily, we don't have to use it. So plugging in what we know we have 4.9 times 10 to the negative 10 equals X squared divided by 0.30. We're gonna cross multiply these two together. When we do that, that's gonna give me X squared equals 1.47 times 10 to the negative 10, we want X not X squared. So take the square root of both sides here. So when I do that, that's gonna give me X which is gonna give me 1.2 times 10 to the negative five molar. Now, what is the question asking me to find? It's asking me to find the concentration of hy hydro ions. When we say that we're really talking about hydro ions at equilibrium. Look at the ice chart here is my hydro ion at equilibrium, it's equal to X which is what I just found. So this number represents the hydro ion concentration at equilibrium. So 1.21 0.2 times 10 of the negative five moer would be my answer.
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example
pH of Weak Acids Example
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7m
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Now, the ph or POH of a weak acid can be calculated once the equilibrium concentration for the hydro num ion is found here, we're going to say that this is determined by using the equilibrium row of the ice chart. Here in this example question, it says what is the ph of a 0.074 molar nitrous acid hno two solution. The K value for the compound is 4.6 times 10 to the negative four. So this K A is less than one. We know that nitrous acid is a weak acid. And because of that, we're gonna have to utilize an ice chart. Now you're going to use steps 1 to 3 to set up the ice chart. So first of all, we're going to have our weak acid react with water, water. He will be in its liquid face said nitrous acid is an acid that means that water will act as a base. Remember based on the brass and Laurie definition, acids donate H plus bases, accept H plus. Doing this would create our nitrite ion plus the hydro lium ion. Here we have our balanced chemical equation and we're gonna say since we're using an ice chart that stands for initial change equilibrium. When it comes to this ice chart, we're gonna plug in the initial amount given for weak acid, which is 0.074 molar. Remember that liquids and solids are ignored within an ice chart. So these would get a, we're not told anything about our products initially. So there's zero for the change grow. Remember we lose reactants to make products. So my react would be minus X products will be plus X step four. Using the equilibrium rule set up the equilibrium expression and solve for X for the equilibrium row, we bring down everything. So this would be 0.074 minus X plus X plus X. Here at this point when we use the equilibrium constant expression, we have to check with our shortcut to see if we can avoid the quadratic formula doing this. We use the 500 approximation method. We're gonna say here when the ratio of the initial concentration divided by RK A. In this case, if it's a ratio greater than 500 we can ignore the minus X within our equilibrium expression. Here. Our initial concentration of the weak acid is 0.074 molar. It's K A is told to us is 4.6 times 10 to the negative four. When we put this into our calculators, we get back 1 60.87. So unfortunately, we don't have a number greater than 500 which means we cannot ignore the minus X. And therefore, we have to set up a quadratic formula. Now, here in this section, we have our equilibrium expression with the values plugged in. But coming back up here, remember that your equilibrium expression is K A equals products over reactants. Again, we ignore solids and liquids K A again is 4.6 times 10 to the negative four at equilibrium. Both our products are X. So multiply together they're X squared divided by 0.074 minus X Ross multiply these together. So we're gonna come over here. So it's gonna be 4.6 times 10 to the minus four. And in parentheses, 0.074 minus X which equals X squared, we're going to distribute, distribute. So when I do that, I'm going to get here uh 3.404 times 10 to the negative five minus 4.6 times 10 to the minus four, X equals X squared. Here are both of our X variables, but the X squared is the one with the larger power. So it's our lead term which means we bring everything over to the left side um actually to the right side to the right side. So we're gonna add 4.6 times 10 to the minus four X to both sides and we're gonna subtract 3.404 times 10 to the negative five from both sides. So when we do this, we're gonna get our equation. Our equation now is going to be X squared plus 4.6 times 10 to the minus four, X minus 3.404 times 10 to the negative five. This would be our A RB and RC. This is important so that we can plug it into the quadratic formula. So now we have negative 4.6 times 10 to the minus four plus or minus the square root of 4.6 times 10 to the negative four squared minus four. A here would just be one, C, don't forget the negative sign divided by two times one. Now, here, when we do the square root of everything within here, we're gonna get X equals negative 4.6 times 10 to the minus four plus or minus 0.011678 divided by two. Now, here it's important to realize that this is plus or minus, which means we're gonna get two answers where X equals negative 4.6 times 10 to the negative four plus 0.011678 divided by two and one where it's negative 4.6 times 10 to negative four minus 0.011678 divided by two. OK. So there's two possibilities. You're either adding or subtracting this number. When we real, when we see that we're gonna get two X values one X will come out to be 0.005609 molar. The other one will come out to be negative 0.006069 molar. Only one of them is the right answer. How do we know which one is the right answer? Well, you would take your X value and no matter where you plug it in terms of the equilibrium row, you should get a positive answer as a result. Because of this, we know that this second one that will not work. Because if I place that negative X here or here, I'd have a negative concentration at equilibrium which is not possible. OK, you cannot have a negative value at equilibrium. So that means the X variable is this value here. And if we look this X variable that we just found gives us H 30 plus that leads us to step by. So for step five, the X variable that we just found gives us the hydro ion concentration. And it can be used to solve for ph remember ph equals negative log of the hydro ion concentration, which again, we said is 0.005609. And then here, when we plug that in, we get 2.25. So 2.25 would be the Ph for nitrous acid solution.
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concept
Calculating Percent Ionization
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1m
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Now, percent ionization or percent association represents the percentage of we of a weak acid that can become ionized when placed in an aqueous solution. Remember that weak acids represent weak electrolytes and strong acids represent strong electrolytes. Because of this weak acids ionize less than 100% when placed in an aqueous solution. And strong acids ionize 100% when placed in an aqueous solution. When we want to calculate the percentage of ionization or dissociation for weak acid, we utilize the following formula and that is the percent ionization or percent association of a weak acid equals the equilibrium concentration of H 30 plus divided by the initial concentration of the weak acid times 100. Now, in order to determine the concentration of hydro ion at equilibrium, they either give it to you or you need to calculate it using an ice chart. So keep this in mind when you need to figure out the percent ionization or percent association of any given weak acid.
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example
pH of Weak Acids Example
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4m
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Calculate the percent association of 4.10 times 10 to the negative one molar of acetic acid. Here, we're told that the K A value for acetic acid is 1.8 times 10 to the negative five. All right, since it's a weak acid, we need to set up an ice chart and we know it's a weak acid because it's K A is less than one. So here we're going to use steps 1 to 3 to set up the ice chart. So first we're going to react the weak acid with water. In this case, water will be in its liquid form. Following the bro and Lori acid definition that we know we know that the acid is acetic acid and therefore the water would be the base, the acid donates an H plus. And the basic subset doing this would create as our products. The acetate ion and the hydro is, this is an ice chart. So which stands for initial change equilibrium. Now we place the initial amount given to us for the weak acid, which we're told is 4.10 times 10 to the negative one, which is just basically 0.410 molar. Remember ice charts ignore solids and liquids. So the water will be ignored. We're not told anything initially about our products. So initially there's zero for the change row. Remember we lose reactants in order to make products. So minus X plus X plus X for the equilibrium row, we bring down everything. So 0.410 minus X plus X plus X. So we filled out our ice chart now using the equilibrium row, set up the equilibrium constant expression and solve for X check if a shortcut can be utilized to avoid the quadratic formula. So here we're going to say with the quadratic formula we have it here. We say we use the 500 approximation method when the ratio of initial concentration to the K A value in this case is greater than 500 we can ignore the minus X. So our initial concentration of acetic acid is 4.1 times 10 to negative one molar. It's K A value was 1.8 times 10 to the negative five. When we punch that in it gives us 22,777 0.8. A number much greater than 500. That means that our, in our equilibrium expression, we can ignore the minus X. So remember our equilibrium expression is products over reactant. So it's K A equals my two products divided by the weak as again, water is a liquid. So we ignore it. So 1.8 times 10 to the negative five equals X squared divided by remember using the 500 approximation method below, we saw that we could ignore the minus X. So this is just 0.410 on the bottom. And we didn't include the minus X here cross multiply these two. When we do that X squared equals 7.38 times 10 to the negative six. Take the scroll to both sides. X equals 2.717 times 10 to the minus three. We just found out what X is and it says use the X variable now to calculate the percent ionization slash association. Here, we're going to say that the X that we found is equal to my hydro ion concentration. So we just found out what H 30 plus concentration is. Remember for weak acid percent ionization or an association equals the concentration of H 30 plus at equilibrium divided by the initial concentration of the acid times 100. So at equilibrium HDO plus was 2.717 at times 10 to the minus three moller, our initial concentration was 0.410 molar. And then multiply that by 100 when we do that, we get 0.66%. So that's the amount of acetic acid that basically ionize when placed in an aqueous solution. It's a weak acid. So we should expect a number that's much smaller than 100%. In this case, in 0.66%.
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Problem
Problem
Calculate the [H+] of a 0.50 M solution of methylammonium bromide, CH3NH3Br.
A
0.50 M
B
1.14 × 10−11 M
C
3.37 × 10−6 M
D
2.97 × 10−9 M
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Problem
Problem
An unknown weak acid has an initial concentration of 0.55 M. What is the pH of the solution if the weak acid also has a pKa of 5.79?
A
0.60
B
6.05
C
3.02
D
5.75
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