Chemistry, 21.04.2020 03:08 chrissy5189

Acetic acid has a Ka of 1.8 * 10-5. Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations: acetic acid ten times greater than acetate, acetate ten times greater than acetic acid, and acetate=acetic acid.
Match each buffer to the expected pH. pH = 3.74 ; pH = 4.74 ; pH = 5.74
Part B: How many grams of dry NH4Cl need to be added to 2.30 L of a 0.600 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.71? Kb for ammonia is 1.8 * 10-5.
Express your answer with the appropriate units.

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Answer from: dogsarecute278

A) acetic acid ten times greater than acetate pH = 3.74

B) acetate ten times greater than acetic acid pH = 5.74

C) For solution 3: acetate=acetic acid pH = 4.74

Mass NH4Cl = 40.30 grams

Explanation:

Step 1: Data given

Ka is: 1.8 * 10^-5

Solution 1: acetic acid ten times greater than acetate

Solution 2: acetate ten times greater than acetic acid

Solution 3: acetate=acetic acid

Step 2: The pH formula

pH = pKa + log[CH3COO-]/[CH3COOH]

For solution 1: acetic acid ten times greater than acetate this means ) [[CH3COO-]/[CH3COOH]) has a value of 1/10

pH = pKa + log[CH3COO-]/[CH3COOH]

pH = -log(1.8*10^-5) + log(1/10)

pH = 4.74 -1

pH = 3.74

For solution 2: acetate ten times greater than acetic acid

pH = pKa + log[CH3COO-]/[CH3COOH]

pH = -log(1.8*10^-5) + log(10)

pH = 4.74 + 1

pH = 5.74

For solution 3: acetate=acetic acid

pH = pKa + log[CH3COO-]/[CH3COOH]

pH = -log(1.8*10^-5) + log(1)

pH = 4.74 + 0

pH = 4.74

Part B: Calculate molarity

pOH = pKb + log [H

5.29= -log (1.8*10^-5) + log [BH+]/[0.600 M]

5.29 = 4.744 + log [BH+]/[0.600 M]

0.546 = [BH+]/0.600

[BH+] = 0.3276 M

Moles NH4+ = 0.3276 M * 2.30 L

Moles NH4+ = 0.75348 moles

Moles NH4Cl = 0.75348 moles

Mass NH4Cl = 0.75348 moles * 53.49 g/mol

Mass NH4Cl = 40.30 grams

Answer from: Quest

$\frac{239}{94}pu+\frac{1}{0}n ====> \frac{y}{z}x+\frac{134}{54}xe+3\frac{1}{0}$

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137 subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54 subtract 54 from both sides.

40 = z

answer

$\frac{103}{40}x=\frac{103}{40}zr$

Answer from: Quest

Answer trigonal pyramidal;

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Acetic acid has a Ka of 1.8 * 10-5. Three acetic acid/acetate buffer solutions, A, B, and C, were ma...

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