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Brrunno [24]
3 years ago
6

Acetic acid has a pKa of 4.74. Buffer A: 0.10 M HC2H3O2, 0.10 M NaC2H3O2 Buffer B: 0.30 M HC2H3O2, 0.30 M NaC2H3O2 Buffer C: 0.5

0 M HC2H3O2, 0.10 M NaC2H3O2 Which of the above buffers has the highest buffer capacity? [ Select ] Which of the above buffers has the lowest buffer capacity? [ Select ]
Chemistry
1 answer:
e-lub [12.9K]3 years ago
6 0

Answer:

Buffer B has the highest buffer capacity.

Buffer C has the lowest buffer capacity.

Explanation:

An effective weak acid-conjugate base buffer should have pH equal to pK_{a} of the weak acid. For buffers with the same pH, higher the concentrations of the components in a buffer, higher will the buffer capacity.

Acetic acid is a weak acid and CH_{3}COO^{-} is the conjugate base So, all the given buffers are weak acid-conjugate base buffers. The pH of these buffers are expressed as (Henderson-Hasselbalch):

pH=pK_{a}(CH_{3}COOH)+log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}

pK_{a}(CH_{3}COOH)=4.74

Buffer A:    pH=4.74+log(\frac{0.10}{0.10})=4.74

Buffer B:   pH=4.74+log(\frac{0.30}{0.30})=4.74

Buffer C:   pH=4.74+log(\frac{0.10}{0.50})=4.04

So, both buffer A and buffer B has same pH value which is also equal to pK_{a} . Buffer B has higher concentrations of the components as compared to buffer A, Hence, buffer B has the highest buffer capacity.

The pH of buffer C is far away from pK_{a} . Therefore, buffer C has the lowest buffer capacity.

     

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a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel
eimsori [14]

Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

  • To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

<u><em>1) What is the partial pressure of SO₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

<u><em>2) What is the partial pressure of N₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

<u><em>3) What is the total pressure in the vessel?</em></u>

  • According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

<em>∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.</em>

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

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Answer:

2. 181.25 K.

3. 0.04 atm.

Explanation:

2. Determination of the temperature.

Number of mole (n) = 2.1 moles

Pressure (P) = 1.25 atm

Volume (V) = 25 L

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) =?

The temperature can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

1.25 × 25 = 2.1 × 0.0821 × T

31.25 = 0.17241 × T

Divide both side by 0.17241

T = 31.25 / 0.17241

T = 181.25 K

Thus, the temperature is 181.25 K.

3. Determination of the pressure.

Number of mole (n) = 10 moles

Volume (V) = 5000 L

Temperature (T) = –10 °C = –10 °C + 273 = 263 K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =?

The pressure can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

P × 5000 = 10 × 0.0821 × 263

P × 5000 = 215.923

Divide both side by 5000

P = 215.923 / 5000

P = 0.04 atm

Thus, the pressure is 0.04 atm

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