Triprotic Acids and Bases Calculations - Video Tutorials & Practice Problems
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1
concept
pH of Weak Triprotic Acid Species
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1m
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When it comes to calculating the pH of weak triprotic species, we need to keep in mind just the acidic form and the basic form. Remember with triprotic species, we have the acidic form where it has possession of all of its H+ ions, all 3 of them. Then we have the 1st intermediate form where it's lost 1, then we have the 2nd intermediate form where it has lost another, and then we finally have the basic form. With triprotic species, we're concerned with the acidic form and the basic form. Now we utilize only Ka 1 to calculate the pH of the acidic form of a weak triprotic acid, And we utilize KB1 to calculate the pH of a basic form of a weak tryprotic acid. Right? So keep in mind we have 4 different forms involved with the tryprotic species. We're mainly concerned with the acidic form and the basic form when it comes to our calculations.
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example
Triprotic Acids and Bases Calculations Example
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6m
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Determine the pH of a 0.225 molar phosphoric acid solution. It will be given 3 ka values since it's triprotic. Now here step 1 is we're going to set up an ice chart for the weak triprotic acid that has it reacting with water. Remember to use the Bronsted Lowry definition to predict the products formed. So here we have our weak triprotic acid reacting with a mole of water or molecule of water. Since it's an acid it's going to donate an H plus to the water. It becomes dihydrogen phosphate. Water gains an h plus, so it becomes h three o plus. We're dealing with an ICE chart since it's a weak acid since its Ka is less than 1, and remember an ICE chart is initial change equilibrium. In an ICE chart we ignore solids and liquids, so the water will be ignored. The initial concentration is 0.225 molar. Our products initially are 0. So remember we've filled out the initial row. For 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, minus x since we're losing this reactant, and plus x since we're making them. We then bring down everything, so 0.225 minuteus x, plus x, plus x. Our ICE chart is completely filled in, so step 4 says using the equilibrium row, set up the equilibrium constant expression with Ka1 and sulfur x. Ka1 is equal to products over reactants, so it equals dihydrogen phosphate ion times the hydronium ion divided by phosphoric acid. Here we place the numbers that we know for Ka 1 in the different concentrations here, so here Ka 1 is 7.5 times 10 to the minus 3, These would be x squared divided by 0.225 minuteus x. Now, once we set up the expression we can check to see if a shortcut can be utilized to avoid the quadratic formula. We do this by utilizing the 500 approximation rule, where we take the initial concentration of our weak triprotic acid, and in this case divided by k a one. If the ratio is greater than 500 we can ignore the minus x within our equilibrium expression. So when we punch this number into our calculator it gives us a number around 30, a number that's much lower than 500. So unfortunately we have to keep the minus x and utilize our quadratic formula here. Alright. So let's go back up and let's do that. We're going to cross multiply these, so we're gonna say 7.5 times 10 to the minus 3, distribute, distribute. When I do that I'm going to get 0.0016875 minus 7.5 times 10 to the minus 3x equals x squared. The x squared is the lead term, everything has to move over to it, so you would add this to both sides and you will subtract this from both sides. Doing that would give us a new equation which is x squared and it's going to be plus 7.5 times 10 to the minus 3x minus 0.0016875. This would be my a, my b, and my c. Setting up the quadratic formula, we'd have negative 7.5 times 10 to the minus 3, plus or minus 7.5 times 10 to the minus 3 squared, minus 4 times 1, negative sign negative.0016875 divided by 2. Solve for this portion here, and so we'd have negative 7.5 times 10 to the minus 3, plus or minus 0.0825 divided by 2. Because of the presence of plus or minus here there's 2 possible answers for x, one where it's plus 0.0825, and one where it's negative or minus 0.0825. So x here could equal 0.0375 or x equals negative 0.045. The correct one, the correct x is the x where you can plug it anywhere in terms of the equilibrium row and your answer is always positive. That means the negative x we cannot use because placing it here or here would give you a negative result. Alright. So we just figured out the correct x to use, which is this one here. That x that we just found, well that x is connected to h3o plus So we could take that x that we just found and find our pH. Alright. So x equals 0.0375, which is equal to the concentration of H3O plus, pH equals negative log of H3O plus, plug it in, and when we do that we're gonna get approximately 1.43 at the pH for this weak triprotic acid. So that would be our final answer.
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Problem
Problem
Determine the pH of 0.250 M sodium phosphate, Na3PO4. Phosphoric acid, H3PO4, contains Ka1 = 7.5 × 10−3, Ka2 = 6.2 × 10−8 and Ka3 = 4.2 × 10−13.
A
12.82
B
7.21
C
1.18
D
12.38
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Problem
Problem
Determine the pOH of 0.450 M citric acid, H3C6H5O7. It possesses Ka1 = 7.4 × 10−4, Ka2 = 1.7 × 10−5 and Ka3 = 4.0 × 10−7.
A
7.602
B
13.653
C
9.230
D
12.260
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