What is the ratio of of [CH 3 COO - ] / [CH 3 COOH] in an acetate buffer at pH = 7? pK a = 4.76.

So this practice problem asks us, what is the ratio of acetate conjugate base over aceticacidconjugate acid in an acetate buffer when the pH is equal to 7. And then it gives us the pKa of acetic acid, which is 4.7 6. And so recall that we're going to need to use the Henderson Hasselbalch equation to determine the ratio of conjugate base over conjugate acid. And so, here is the, Henderson Hasselbalch equation. So what we can do is start to plug in our given values. And so we're given that the pH is equal to 7, as well as the pKa is equal to 4.76. So let's go ahead and start plugging in our values. So the pH is equal to 7, the pKa is equal to 4.76, and then we still have plus log of conjugate base, which I'll abbreviate as c b here, over conjugate acid. And so when we subtract 4.76 from both sides of the equation, what we get is 2.24 is equal to log of the concentration conjugate base over the concentration of conjugate acid. And then at this point, what we need to do is get rid of the log and recall that our logarithm properties, the way that we get rid of the log, is to do the antilog. And all the antilog is is taking log which has a base 10, so taking 10, and raising it to the 2.24. So when we take 10 and raise it to the exponent of 2.24, that gets rid of our log. And so all we're left with is the concentration of conjugate base over the concentration of conjugate acid. And so when you take 10 and you raise it to the 2.4, the answer actually comes out to 173.78001. 1. And of course, this is right about equal to a 173.78, which is one of our answer options and that is b. And so the other answer options are incorrect and we can scratch them off, and of course that leaves b here as our correct answer, and I'll see you guys in our next practice video.

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2. Water

Henderson Hasselbalch Equation

Biochemistry

2. Water

Henderson Hasselbalch Equation

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Multiple Choice

What is the ratio of of [CH ₃COO^-] / [CH₃COOH] in an acetate buffer at pH = 7? pK _a = 4.76.

122.43

173.78

39.84

96.31

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Verified step by step guidance

Identify the relevant equation for buffer solutions, which is the Henderson-Hasselbalch equation: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \), where [A^-] is the concentration of the conjugate base and [HA] is the concentration of the acid.

Substitute the given values into the Henderson-Hasselbalch equation. Here, \( \text{pH} = 7 \) and \( \text{pK}_a = 4.76 \).

Rearrange the equation to solve for the ratio \( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \): \( 7 = 4.76 + \log \left( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \right) \).

Subtract \( 4.76 \) from both sides to isolate the logarithmic term: \( 7 - 4.76 = \log \left( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \right) \).

Convert the logarithmic expression to its exponential form to find the ratio: \( 10^{(7 - 4.76)} = \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \).