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 trip protic species, we need to keep in mind just the acidic form and the basic form, remember with tripod species, we have the acidic form where it has position possession of all of its H plus ions, all three of them. Then we have the first intermediate form where it's lost one. Then we have the second intermediate format has lost another. And then we finally have the basic form with trip protic species. We're concerned with the acidic form and the basic form. Now we utilize only K A one to calculate the ph of the acidic form of a weak trip protic acid. And we utilize KB one to calculate the ph of a basic form of a weak Tripodi acid, right? So keep in mind, we have four different forms involved with the trip protic species. More 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. They were given three K A values since its trip product. Now here's step one is we're going to set up an ice chart for the weak tr protic acid that has it reacting with water. Remember to use the Bronson or definition to predict the products formed. So here we have our weak tr protic acid reacting with a mole of water or molecule of water. Since it's an acid, it's gonna donate an H plus the water, it becomes di hydrogen phosphate, water gained in H plus. So becomes H +30 plus. We're dealing with an ice chart since it's a weak acid, since it's K A is less than one. And remember an ice chart is initial change equilibrium. In a nice chart, we ignore solid, the liquids. So the water will be ignored. The initial concentration is 0.225 molar, our products initially are zero. So remember we filled out the initial row for 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 minus accents were losing this reactant and plus accents were making them, we then bring down everything. So 0.225 minus X plus X plus X, our ice chart is completely filled in. So step four says using the equilibrium row, set up the equilibrium constant expression with K A one and sulfur X K A one is equal to products over reactants. So it equals di hydrogen phosphate ion times the hydro ion divided by phosph forecast. Here we place the numbers that we know for K A one in the different concentrations here. So here K A one is 7.5 times 10 to the minus three. These would be X squared divided by 0.225 minus X. Now, once we set up the expression, we u, 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 trip protic 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. All right. 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 three, distribute, distribute. When I do that, I'm going to get 0.0016875 minus 7.5 times 10 to the minus three X equals X squared. The X squared is the lead term. So 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 gonna be plus 7.5 times 10 to the minus three, X 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 three plus or minus 7.5 times 10 to the minus three squared minus four times one negative sign negative 0.0016875 divided by two solve for this portion here. And so we'd have negative 7.5 times 10 to the minus three plus or minus 0.0825 divided by two because of the presence of plus or minus here. There's two 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 a negative X, we cannot use because placing it here or here would give me a negative result. All right. 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 H 30 plus. So we could take that A, that we just found and find our PH. All right. So X equals 0.0375 which is equal to the concentration of H 30 plus PH equals negative log of H 30 plus. Plug it in. And when we do that, we're gonna get approximately 1.43 at the PH for this weak trip protic 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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