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. Here we'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 bases, 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 we're tested calculate the hydroxide ion concentration of a 0.55 molar potassium fluoride solution at 25 degrees Celsius. Here we're told the acid dissociation constant of h f, hydrofluoric acid is 3.5 times 10 to the negative 4. Alright. So step 1 is we're gonna set up an ICE chart for the weak base that has a reacting with water. Now we know that this is a weak base 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 3 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 to 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. Alright, so F- here is our weak base. Now for ionic bases which this one is, we ignore the neutral metal cation, here we ignore the potassium ion. Now, use the Bronsonory definition to predict the products form, Make sure that Kb is used in the presence of the weak base. So right now, we are only given the Ka of its acid form, but we're gonna need to utilize its Kb later on to find our final answer. So remember, we're gonna 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+ over to F minus. This would create h f aqueous plus Oh minus aqueous as products. Step 2, use the initial row. So, remember for this I stands for initial change equilibrium. Using the initial row, place the amount given for the weak base. So the weak base we're told is 0.55 molar. Place a 0 for any substance not given an initial amount. In an ICE chart we ignore solids and liquids, so the water is ignored. We're not told anything about our products initially so they're both 0. Step 3, 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 row, set up the equilibrium constant expression with a b, again because it's a weak base, and solve for x. We checked this if a shortcut can be utilized to avoid the quadratic formula. Alright. 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 k w equals k a times k b. We need to isolate kb, so kb equalskw divided by ka. So this would be 1.0 times 10 to the negative 14, divided by our ka which is 3.5 times 10 to the negative 4. Doing that gives us a kv value of 2.857 times 10 to the negative 11. Alright. So our initial concentration is 0.55 molar, we found our kb as 2.85 7 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 a liquid so we ignore it. So bringing down this expression and inserting values for what we have, AB we said is 2.857 times 10 to the negative 11. Both the products are X at equilibrium, so multiplied together is x squared, and this will be 0.55 minus x. Well, remember we just did the ratio of initial concentration divided by k b, 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 going to do that up here, so when I do that I'm going to get x squared equals 1.57135 times 10 to the negative 11. I don't want x squared, I just want x, so take the square root of both sides. X equals 3.96 times 10 to the negative 6 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 find out what my x value is, which is equivalent to my hydroxide ion concentration. So So my final answer would be 3.96 times 10 to the negative 6 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 kV value of ethyl amine is 5.6 times 10 to the negative 4. Alright, so we're going to start out by following the steps 1 to 3 and setting up an ICE chart. So here we have our ethyl amine. The ethyl amine being a weak base because its kV is less than 1, would react with liquid water. Since it's the base, water will act as the acid donating an H+ to it. This H+ will go towards the nitrogen giving us C2H5 NH3 positive. This is the ethyl ammonium ion. Water 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 row we're going to place the initial amount given forethylamide, which is 0.12 molar. Liquids and solids are ignored within a nice charge, so water will be ignored. Our products initially are 0 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 minuteus x, plus x, plus x. This leads us to step 4, we're using the equilibrium row set up the equilibrium constant expression and solve for X. Here our equilibrium constant expression, since this is a base we have kB, equals products over reactants, so it equals ethylammonium ion times hydroxide ion over ethyl amine. 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, to our kv 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 again remember is negative b plus or minus square root of b squared minus 4ac over 2a. So come up here, our expression is 5.6 times 10 to the minus 4 for the weak base, and that'd be x squared over 0.12 minuteus x. Cross multiply those, so when we do that we're gonna get 5.6 times 10 to the -4, and in parenthesis 0.12 x, and that equals x2. Distribute, distribute. So that give me 5.6 times oops, 5.6 times negative 4 times 10 to the negative 4 times 0.12 gives me 6.72 times 10 to the negative 5 minus 5.6 times 10 to the minus 4x, and this equals x squared. Now my 2x 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 4 x to both sides, and you'd subtract 6.72 times 10 to the negative 5, on both sides. When we do that we have our new equation, so x squared plus 5.6 times 10 to the negative 4x minus 6.72 times 10 to the minus 5. Here this equation would equal a, b, and c. Setting up our quadratic formula we have negative 5.6 times 10 to the minus 4, plus or minus the square root of 5.6 times 10 to the minus 4 squared minus 4 times, there's a one here, one times negative, don't forget the negative sign, negative 6.72 times 10 to the minus 5, divided by 2 times 1. 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 4, plus or minus 0.0164 divided by 2. So here there are 2 outcomes for x because it's plus or minus, so x here could equal negative 5.6 times 10 to the minus 4 plus 0.0164 divided by 2, or x could equal negative 5.6 times 10 to the minus 4, minus 0.0164 divided by 2. This gives us 2 answers for x initially. So in the first top one we get 0.00792, the molar, and in this bottom one we get negative 0.00848 molar. But which one is the correct one? The correct X is the X that where wherever you place it, in terms of the equilibrium row, 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. Alright, so here the x that we've just found takes us to step 5. The x variable that we just found would give us our hydroxide ion concentration. Since we know hydroxide ion concentration we can 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 pOH. And then remember that pH equals negative log of H+ and that would give me my pH. Or you could also remember that pH +0h equals 14. So pH equals 14 minuteus pOH. So here would be 14 minuteus 2.10, and if it's 2.10 that would come out to 11.90. So we'd have 11.90 as our pH for this solution. Okay, so just remember when we find x look on the equilibrium row. That x will give you either H3O 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 dissociation, 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, we'll 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 grams of sodium hypoiodide are dissolved with 500 ml of solution. We're told that the K value of hypoiodis acid, which is HIO is 2.3 times 10 to the negative 11. Alright. So, we're going to say that hypoiodic acid is a weak acid since its Ka is less than 1. 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 cation, in this case the sodium ion. This leaves us with the hypoiodide 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+0H- as products. We're dealing with the weak base, so this is our ice chart setup. 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 grams of sodium iodide, we're gonna convert that into moles. So we're gonna say here for every 1 mole of sodium iodide we have a mass of 165.89 grams of sodium hypoiodide. That's the mass of the sodium, the iodine, and oxygen together. So here, the grads cancel out, and we're gonna have here for every 1 mole we have of sodium iodide, we have 1 mole of iodide ion, which equals 0.441 moles. We need molarity for the ICE chart, so molarity here will be moles over liters, so it would be 0.441 moles, divided by 0.500 liters, 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 they're 0. We lose reactants in order to make product, Bring down everything. Now here using the equilibrium row, 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. Kb would equal products over reactants. However, we're not given Kb initial we're given Ka, so we have to remember that Ka times Kb equals Kw. So we're gonna say Kb here equals kw divided by ka, so here this would be 1.0x10-fourteen, divided by our k a which we're told is 2.3 times 10 to the negative 11. Doing this gives us our kv value, our kv value here would equal 4.3 times 10 to the negative 4. Now that we know the kv value we can do the 500 approximation method. When the ratio of the initial concentration divided by in this case kv, 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 kv is 4.3 times 10 to the negative 4. When we punch that in that gives us 2,051.2. 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 x we can avoid the quadratic formula. We're going to cross multiply these 2. So when we do that we get x squared equals 3.7 times 10 to the minus 4, and then take the square root of both sides. X will give me approximately 0.01-947. When we find x here, x gives me Oh- concentration. So that takes us to step 5, we would use the x variable which we just found to help us calculate the percent ionization or dissociation. So remember, here we're going to say percent ionization or percent dissociation for a weak base equals the 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 will represent the percent ionization of my hypoiodide ion.
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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.