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blsea [12.9K]
3 years ago
11

What is the pH of 10.0 mL solution of 0.75 M acetate after adding 5.0 mL of 0.10 M HCl (assume a Ka of acetic acid of 1.78x10-5)

Chemistry
2 answers:
ValentinkaMS [17]3 years ago
4 0

Answer:

5.90

Explanation:

Initial moles of CH3COO- = 10.0/1000 x 0.75 = 0.0075 mol

Moles of HCl added = 5.0/1000 x 0.10 = 0.0005 mol

CH3COO- + HCl => CH3COOH + Cl-

Moles of CH3COO- left = 0.0075 - 0.0005 = 0.007 mol

Moles of CH3COOH formed = moles of HCl added = 0.0005 mol

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

= -log Ka + log(moles of CH3COO-/moles of CH3COOH)

= -log(1.78 x 10^(-5)) + log(0.007/0.0005)

= 5.90

Svetach [21]3 years ago
4 0

Answer:

The correct answer is 5.895.

Explanation:

The reaction will be,

CHCOO⁻ + H+ ⇔ CH₃COOH

Both the HCl and the acetate are having one n factor.

The millimoles of CH₃COO⁻ is,

= Volume in ml × molarity = 10 × 0.75 = 7.5

The millimoles of HCl = Volume in ml × molarity = 5 × 0.1 = 0.5

Therefore, 0.5 will be the millimoles of CH₃COOH formed, now the millimoles of the CH₃COO⁻ left will be, 7.5-0.5 = 7.0

The volume of the solution is, 10+5 = 15 ml

The molarity of CH₃COO⁻ is, millimoles / volume in ml = 7/15

The molarity of CH₃COOH is 0.5/15

pH = pKa + log[CH₃COO⁻]/[CH₃COOH]

= 4.74957 + 1.146

= 5.895

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Answer:

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Explanation:

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8 0
2 years ago
What is the ph of a solution which is 0. 023 m in weak base and 0. 037 m in the conjugate weak acid ( a = 7. 1 × 10^−6)?
Rzqust [24]

The pH of a solution which is 0.023 m in weak base and 0 037 m in the conjugate weak acid whose Ka= 7.1 × 10⁻⁶ is 4.93.

pH determines the amount of hydrogen and hydroxide ions in a solution. It is the negative logarithm of hydrogen ion concentration.

Given,

Ka= 7.1 × 10⁻⁶

Weak base = 0.023M

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Using base dissociation constant, Ka we can calculate pKa by:

pKa = -log [Ka]

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A weak base and its conjugate acid are present in equal proportions in buffer solution. Using the Henderson-Hasselbalch equation, we can find out the pH of a buffer solution that constitutes a weak base and its conjugate acid.

Using Henderson-Hasselbalch equation,

pH = pKa + log [Weak base / acid]

pH = 5.15 + log [0.023 / 0.037]

pH = 5.15 + log[0.6]

pH = 5.15 - 0.22

pH = 4.93

Therefore, the pH of the solution is 4.93.

Learn more about pH here, brainly.com/question/22390063

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ki77a [65]

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