Henderson-Hasselbalch Equation - Video Tutorials & Practice Problems
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1
concept
Henderson-Hasselbalch Equation
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Now, the Henderson Hasselback equation allows you to calculate the PH of a buffer without having to use an ice chart. Here, we're going to say it only applies to buffers composed of a conjugate acid base fare. We can look at the Henderson Hasselback equation as two different formulas. The formula that you use is based on if they give you the K A or KB of your buffer solution. If they give you the K A of your buffer solution, you can say that PH equals PK A plus log of conjugate base over weak acid. Now, if they give you PK, if they give you KB, then you could use this formula PH equals PKB plus log of conjugate acid over weak base. Now, when it comes to the sign the brackets, we know that it tends to mean molarity or concentration for the Henderson Hasselback equations. It could also be used for molts. OK. So just remember the units that can go within these brackets can be either molarity or moles. And remember that moles itself equals liters times molarity. So keep that in mind when you utilize the Henderson Hasselback equations.
2
example
Henderson-Hasselbalch Equation Example
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Calculate the ph of a solution containing 2.0 molar of nitrous acid and 1.48 molar of lithium nitrate. Here, we're told that the K A of our weak acid is 4.6 times 10 to the negative four. Now here, because the K A value is less than one. We know that nitrous acid is a weak acid. Lithium nitrite looks similar to nitrous acid except it has one less H plus ion because of this, this has to be the conjugate base. So we have a weak acid, we have a conjugate base. This is the pairing that helps to make a buffer. So we're gonna use the Henderson Hasselback equation to calculate the ph of this buffer. We're gonna say Ph equals now because they give us K A, we can say ph equals PK A plus log of conjugate base over a weak acid. Here PK A remembers just negative log of K A. So negative log of 4.6 times 10 to the minus four plus log of conjugate base amount, which is gonna be 1.48 molar divided by weak acid amount, which is 2.0 mole. When we plug this in, we get 3.21 as the PH for this buffer solution.
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Problem
Problem
The Kb of C6H5NH2 (aniline) is 3.9 × 10−10. Determine pH of a buffer solution made up of 500 mL of 1.4 M C6H5NH2 and 230 mL of 2.3 M C6H5NH3+.
A
4.81
B
9.62
C
4.38
D
9.29
4
Problem
Problem
Determine the buffer component concentration ratio (CB/WA) for a buffer with a pH of 4.7. Ka of boric acid (H3BO3) is 5.4 × 10−10.
A
4.568 : 1
B
2.706 × 10−5 : 1
C
1 : 4.568
D
1 : 2.706 × 10−5
5
Problem
Problem
Calculate mass of NaN3 that needs be added to 1.8 L of 0.35 M HN3 in order to make a buffer with a pH of 6.5. Ka of hydrazoic acid is 1.9 × 10−5.
A
1.4 g NaN3
B
1.0 × 10−2 g NaN3
C
2.5 × 103 g NaN3
D
3.55 g NaN3
6
concept
Calculating Buffer Range
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Now, when it comes to calculating buffer range, first, it's important to remember that buffers are effective at a specific Ph range. To figure that out, we see that Ph equals PK A plus or minus one. That's the range, the PH range in which a buff will act most effectively at resisting a sharp change in ph. Now recall that a buffer is ideal though when the concentration of weak acid is equal to the concentration of conjugate base or you could say when the concentration of weak base is equal to the concentration of conjugate acid. Same thing. Now this is because the Ph of the buffer will be equal to the PK A of the weak acid and this will resist a Ph change the best. Now here, if we had an example, we have PH equals PK A plus log gov conjugate base over weak acid. Remember when it comes to the Henderson Hasselback equation, we can observe it two different ways. If they're giving you K A, you can use the top version where PH equals PK A plus lot of conjugate base over weak acid. The bottom one you use, if they give you KB. Here, Ph equals PKB plus laga conjugate acid of a weak base. Well, going back to this, we can see that both the conjugate base and weak acid amounts are equal to one another. So 0.40 divided by 0.40 is just equal to one. And remember if we're dealing with log of one, which this is if you punch that into your calculator log of one gives you zero. So the equation simplifies to Ph equals PK A plus zero. And if you drop out the +02, then Ph equals PK A. So when the amount of conjugate base and weak acid are equal to each other, or when the amount of conjugate acid is equal to weak base, we have an ideal buffer. And then the equation, the Henderson Hasselback equations simplified down to PH equals PK A or PH equals PKB depending on which one you're using, right. So keep that in mind, the effectiveness of a buffer happens best within a PH range of Ph equals PK A plus or minus one. And it's most ideal when the concentrations of the species are equal to one of them.
7
example
Henderson-Hasselbalch Equation Example
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1m
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Here in this example, it says determine the buffering range of a solution containing lactic acid which has a K of 1.4 times 10 to the negative four and sodium lactate its conjugate base. Now, here we're looking for a buffering range. Remember that when it comes to your buffer range, the PH range, so the range in which the buffer works most effectively is PH equals PK A plus or minus one. Remember that PK A is equal to negative log of K A. So just take the negative log of this K A value. When we do that, we get 3.85. So that means our buffering range or PH range for the effectiveness of a buffer is equal to 3.85 plus or minus one. That would mean our range is 3.85 minus 123.85 plus one, which would translate to a range of 2.85 ph two 4.85 ph. OK. This would be our buffering range in which this particular buffer will be most effective.
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Problem
Problem
Which of the following weak acid-conjugate base combinations would result in an ideal buffer solution with a pH of 9.4?
a) formic acid (HCHO2) and sodium formate (Ka = 1.8 x 10-4)
b) benzoic acid (HC7H5O2) and potassium benzoate (Ka = 6.5 x 10-5)
c) hydrocyanic acid (HCN) and lithium cyanide (Ka = 4.9 x 10-10)
d) iodic acid (HIO3) and sodium iodate (Ka = 1.7 x 10-1)
A
formic acid (HCHO2) and sodium formate (Ka = 1.8 x 10-4)
B
benzoic acid (HC7H5O2) and potassium benzoate (Ka = 6.5 x 10-5)
C
hydrocyanic acid (HCN) and lithium cyanide (Ka = 4.9 x 10-10)
D
iodic acid (HIO3) and sodium iodate (Ka = 1.7 x 10-4)
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