So I have the correc answer **Design a Buffer with a pH of 6.94**To design a buffer with a pH of 6.94, one can use the weak base/conjugate acid systems listed below:| Weak Base | \(K_b\) | Conjugate Acid | \(K_a\) | \(pK_a\) ||------------|---------------|----------------|-------------|----------|| \(\text{CH}_3\text{NH}_2\) | \(4.2 \times 10^{-4}\) | \(\text{CH}_3\text{NH}_3^+\) | \(2.4 \times 10^{-11}\) | 10.62 || \(\text{C}_6\text{H}_5\text{O}_3\text{N}\) | \(5.9 \times 10^{-7}\) | \(\text{C}_6\text{H}_5\text{O}_3\text{NH}^+\) | \(1.7 \times 10^{-8}\) | 7.77 || \(\text{C}_5\text{H}_5\text{N}\) | \(1.5 \times 10^{-9}\) | \(\text{C}_5\text{H}_5\text{NH}^+\) | \(6.7 \times 10^{-6}\) | 5.17 |**Calculation Task:**Determine how many grams of the bromide salt of the conjugate acid must be combined with how many grams of the weak base to produce 1.00 L of a buffer that is 1.00 M in the weak base.- Grams of bromide salt of conjugate acid: **150.2**- Grams of weak base: **20.96**

The amount of weak acid and it's salt required can be calculated using the Henderson-hasselbalch equation.Answer: Here we need to prepare buffer using weak base and bromide salt of its conjugate base. Given: [weak base] = 1.00 M pH required = 6.94 => pOH = 14 - pH = 14 - 6.94 = 7.06 Volume of buffer solution = 1.00 L So we will choose the pair whose pKa is near to 6.94 to get high buffer capacity. Since the nearest pKa is 7.77 => our pair of weak base and it's bromide salt is C6H15O3N and C6H15O3NHBr respectively. given Kb for weak base = 5.9 × 10-7 => pKb = -log(5.9 × 10-7 ) = 6.2291 Now we will find concentration of salt: Applying Henderson hasselbalch equation: pOH = pKb + log([weak base] / [bromide salt]) => 7.06 = 6.2291 + log(1.00 M / [bromide salt]) => log(1.00 M / [bromide salt]) = 7.06 - 6.2291 => log(1.00 M / [bromide salt]) = 0.8309 => (1.00 M / [bromide salt]) = 6.7749 => [bromide salt] = 1.00 M / 6.7749 => [bromide salt] = 0.1476 M Now calculation for amounts: Bromide salt is C6H15O3NHBr molar mass = 230.1001 g/mol [C6H15O3NHBr] = 0.1476 M Volume of buffer given = 1.00 L => Moles of C6H15O3NHBr = M × V = 0.1476 M × 1.00 L = 0.1476 mol => mass of bromide salt C6H15O3NHBr = moles × molar mass = 0.1476 mol × 230.1001 g/mol = 33.9636 Molar mass of weak base C6H15O3N = 149.1882 g/mol => Moles of weak base C6H15O3N = M × V = 1.00 M × 1.00 L = 1.00 mol => Mass of weak base C6H15O3N = moles × molar mass = 1.00 mol × 149.1882 g/mol = 149.1882 g So the answers are: Grams of bromide salt of conjugate acid = 33.9636 g Grams of weak base = 149.1882 g Thank you!

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Chemistry

Design a buffer that has a pH of 6.94 using one of the weak base/conjugate acid systems shown below. Weak Base Conjugate Acid Ka pKa CH;NH, 4.2x10-4 CH;NH; 2.4x10-11 10.62 CH1503NH C3H3NH CGH1503N 5.9x10-7 1.7x108 7.77 C;H;N 1.5x109 6.7x10-6 5.17 How many grams of the bromide salt of the conjugate acid must be combined with how many grams of the weak base, to produce 1.00 L of a buffer that is 1.00 M in the weak base? grams bromide salt of conjugate acid = 150.2 grams weak base 20.98

Design a buffer that has a pH of 6.94 using one of the weak base/conjugate acid systems shown below. Weak Base Conjugate Acid Ka pKa CH;NH, 4.2x10-4 CH;NH; 2.4x10-11 10.62 CH1503NH C3H3NH CGH1503N 5.9x10-7 1.7x108 7.77 C;H;N 1.5x109 6.7x10-6 5.17 How many grams of the bromide salt of the conjugate acid must be combined with how many grams of the weak base, to produce 1.00 L of a buffer that is 1.00 M in the weak base? grams bromide salt of conjugate acid = 150.2 grams weak base 20.98

Chemistry

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

Question

So I have the correc answer

**Design a Buffer with a pH of 6.94**

To design a buffer with a pH of 6.94, one can use the weak base/conjugate acid systems listed below:

| Weak Base | \(K_b\) | Conjugate Acid | \(K_a\) | \(pK_a\) |
|------------|---------------|----------------|-------------|----------|
| \(\text{CH}_3\text{NH}_2\) | \(4.2 \times 10^{-4}\) | \(\text{CH}_3\text{NH}_3^+\) | \(2.4 \times 10^{-11}\) | 10.62 |
| \(\text{C}_6\text{H}_5\text{O}_3\text{N}\) | \(5.9 \times 10^{-7}\) | \(\text{C}_6\text{H}_5\text{O}_3\text{NH}^+\) | \(1.7 \times 10^{-8}\) | 7.77 |
| \(\text{C}_5\text{H}_5\text{N}\) | \(1.5 \times 10^{-9}\) | \(\text{C}_5\text{H}_5\text{NH}^+\) | \(6.7 \times 10^{-6}\) | 5.17 |

**Calculation Task:**

Determine how many grams of the bromide salt of the conjugate acid must be combined with how many grams of the weak base to produce 1.00 L of a buffer that is 1.00 M in the weak base.

- Grams of bromide salt of conjugate acid: **150.2**
- Grams of weak base: **20.96**

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