Determine how many grams of sodium acetate, NaCH 3 CO 2 (MW:82.05 g/mol), you would mix into enough 0.065 M acetic acid CH 3 CO 2 H (MW:60.05 g/mol) to prepare 3.2 L of a buffer with a pH of 4.58. The K a is 1.8 x 10 -5 .

Hey, guys, we're not gonna attempt to do the practice question that was left on the bottom of the page, so let's take a look at it. So here it says, determine how many grams of sodium acetate you would mix in tow. Enough 0.65 Mueller acetic acid to prepare 3. leaders of a buffer with a pH of 4.58 Now, since I use the word buffer, we know we have to use the buffer equation. And since we're gonna need room to work this out, guys, I'm gonna remove myself from the image so we have more room to work with. Alright, so let's take a look. We know we're dealing with Henderson hassle back, so ph equals PK plus log of conjugate base over weak acid. PH is 4.58 Then we have 4.744 We got this when we took the negative log of the K A here. So that's plus log off. We don't know anything about the conjugate base, so we're gonna say this is X. But we do know the mole arat e off the weak acid. Yeah, all we gotta do now is salt for that x variable. If we know acts will know the polarity of our conjugate base. And from there we can work it out to get 2 g. So we're gonna do next is we're going to subtract 4.74473 from both sides. So we're gonna get negative 0.1647 to equals log off X over 0. Mueller. Now we need to break into that into those brackets to get to the X. But before we could do that, we have to get rid of that log. So we're gonna divide both sides by log, which is like taking the anti log. Remember, when we're dividing something by law, he becomes 10 to that number and noticed it was negative. So it stays negative. The sign does not change. It stays negative because it began as negative. And that will equal X over 065 Moeller multiply both sides by 0.65 Moeller um so my ex here will give me the molar ity of my conjugate base, which is 0.4482 Moeller n a C H three co two. Now remember that moles equals leaders times molar ity Because remember, morality is moles over leaders. So if they multiply, my leaders were cancel out. And that's how I end up with moles. We have the polarity of our conjugate base. We have the leaders of the solution. We can multiply them together to find our moles. So we're gonna take the 3.2 leaders multiplied by the 0.4482 Moeller of my conjugate base, and you will now know your moles off conjugate base. And now that we know the moles, I gave you its molecular weight earlier. So we're gonna say, for every one mole of it, we have 82.5 g, so you'll get 11.68 g of your conjugate base. Okay, So the key was this to this question, was first realizing you have ah, weak acid and its conjugate base. That's what's most important because they don't necessarily have to tell you have a buffer. You have to be able to see that you have a buffer. You have a weak acid conjugate base means you have one less H So we use the buffer equation toe isolate our X, which in this case would give us our modularity of the conjugate base. Multiply it Times leaders to get moles, change moles to grams. Following those steps were the key to finding out the answer for this question.

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18. Aqueous Equilibrium

Intro to Buffers

General Chemistry

18. Aqueous Equilibrium

Intro to Buffers

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Multiple Choice

Determine how many grams of sodium acetate, NaCH₃CO₂ (MW:82.05 g/mol), you would mix into enough 0.065 M acetic acid CH₃CO₂H (MW:60.05 g/mol) to prepare 3.2 L of a buffer with a pH of 4.58. The K_a is 1.8 x 10^-5.

10.90 grams

11.68 grams

6.35 grams

2.21 grams

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Verified step by step guidance

Identify the components of the buffer system: acetic acid (CH3CO2H) and sodium acetate (NaCH3CO2). The buffer system is based on the weak acid and its conjugate base.

Use the Henderson-Hasselbalch equation to relate the pH of the buffer to the concentrations of the acid and its conjugate base: \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Calculate the pKa from the given Ka: \( \text{pKa} = -\log(1.8 \times 10^{-5}) \).

Rearrange the Henderson-Hasselbalch equation to solve for the ratio \( \frac{[\text{A}^-]}{[\text{HA}]} \): \( \frac{[\text{A}^-]}{[\text{HA}]} = 10^{(\text{pH} - \text{pKa})} \). Substitute the given pH and calculated pKa to find this ratio.

Calculate the moles of acetic acid (CH3CO2H) in the solution: \( \text{moles of CH3CO2H} = \text{Molarity} \times \text{Volume} = 0.065 \text{ M} \times 3.2 \text{ L} \).

Using the ratio from step 3 and the moles of acetic acid from step 4, calculate the moles of sodium acetate (NaCH3CO2) needed. Convert these moles to grams using the molar mass of sodium acetate: \( \text{grams of NaCH3CO2} = \text{moles of NaCH3CO2} \times 82.05 \text{ g/mol} \).