Now, the Henderson-Hasselbalch 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 pair. We can look at the Henderson-Hasselbalch equation as two different formulas. The formula that you use is based on if they give you the Ka or Kb of your buffer solution. If they give you the Ka of your buffer solution, you can say that pH equals pKa plus logconjugate baseweak acid. Now, if they give you Kb, then you could use this formula: pH equals pKb plus logconjugate acidweak base. Now, when it comes to the sign, the brackets, we know that it tends to mean molarity or concentration. For the Henderson-Hasselbalch equations, it could also be used for moles. Okay. So, just remember that the units that can go within these brackets can be either molarity or moles. Remember that moles itself equals liters times molarity, so keep that in mind when you utilize the Henderson-Hasselbalch equations.
Henderson-Hasselbalch Equation - Online Tutor, Practice Problems & Exam Prep
Henderson-Hasselbalch Equation
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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 Ka of our weak acid is 4.6 × 10 - 4 . Now here because the Ka value is less than 1, we know that nitrous acid is a weak acid. Lithium nitrate looks similar to nitrous acid, except it has one less H+ ion. Because of this, this has to be the conjugate base. So, we have a weak acid, and we have a conjugate base. This is the pairing that helps to make a buffer. So we're going to use the Henderson-Hasselbalch equation to calculate the pH of this buffer. We're going to say pH equals now because they give us Ka, we can say pH equals pKa plus log of conjugate base over weak acid. Here pKa, remember, is just negative log of Ka. So, negative log of 4.6 × 10 - 4 , plus log of conjugate base amount, which is going to be 1.48 molar, divided by weak acid amount which is 2.0 molar. When we plug this in, we get 3.21 as the pH for this buffer solution.
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+.
4.81
9.62
4.38
9.29
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.
4.568 : 1
2.706 × 10−5 : 1
1 : 4.568
1 : 2.706 × 10−5
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.
1.4 g NaN3
1.0 × 10−2 g NaN3
2.5 × 103 g NaN3
3.55 g NaN3
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 say that pH=pKa±1. That's the range, the pH range, in which a buffer will act most effectively in resisting a sharp change in pH.
Now recall that a buffer is ideal 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. This is because the pH of the buffer will be equal to the pKa of the weak acid, and this will resist a pH change the best.
Now here, if we had an example, we have pH=pKa+log(conjugatebaseweakacid). Remember, when it comes to the Henderson-Hasselbalch equation we can observe it in 2 different ways. If they're giving you Ka, you can use the top version where pH=pKa+log(conjugatebase/weakacid). The bottom one you use if they give you Kb. Here pH=pKb+log(conjugateacid/weakbase). 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 1. And remember, if we're dealing with log of 1, which this is, you punch that into your calculator, log of 1 gives you 0. So the equation simplifies to pH=pKa+0. And if you drop off the 0, then pH=pKa. 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 Henderson-Hasselbalch equations simplify down to pH=pKa or pH=pKb, depending on which one you're using.
Alright? So keep that in mind. The effectiveness of a buffer happens best within a pH range of pH=pKa±1, and it's most ideal when the concentrations of the species are equal to one another.
Henderson-Hasselbalch Equation Example
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Here in this example it says, determine the buffering range of a solution containing lactic acid, which has a Ka1.4 times 10-4, 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=pKa±1. Remember that pKa=-logKa. So just take the negative log of this Ka value. When we do that, we have 3.85. So, that means our buffering range or pH range for the effectiveness of a buffer is equal to 3.85±1. That would mean our range is 3.85-1to3.85+1, which will translate to a range of 2.85pHto4.85pH. Okay. This would be our buffering range in which this particular buffer will be most effective.
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)
formic acid (HCHO2) and sodium formate (Ka = 1.8 x 10-4)
benzoic acid (HC7H5O2) and potassium benzoate (Ka = 6.5 x 10-5)
hydrocyanic acid (HCN) and lithium cyanide (Ka = 4.9 x 10-10)
iodic acid (HIO3) and sodium iodate (Ka = 1.7 x 10-4)
