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Lostsunrise [7]
3 years ago
11

You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of a

cetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution?
Chemistry
1 answer:
balandron [24]3 years ago
6 0

Answer:

56.9 mmoles of acetate are required in this buffer

Explanation:

To solve this, we can think in the Henderson Hasselbach equation:

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

To make the buffer we know:

CH₃COOH  +  H₂O  ⇄   CH₃COO⁻  +  H₃O⁺     Ka

We know that Ka from acetic acid is: 1.8×10⁻⁵

pKa = - log Ka

pKa = 4.74

We replace data:

5.5 = 4.74 + log ([acetate] / 10 mmol)

5.5 - 4.74 = log ([acetate] / 10 mmol)

0.755 = log ([acetate] / 10 mmol)

10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)

5.69 = ([acetate] / 10 mmol)

5.69 . 10 = [acetate] → 56.9 mmoles

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Which solution below has the highest concentration of hydroxide ions?a. pH= 3.21b. pH= 7.00c. pH= 7.93d. pH= 12.59e. pH= 9.82
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Answer:

  • <em>The solution that has the highest concentration of hydroxide ions is </em><u>d. pH = 12.59.</u>

Explanation:

You can solve this question using just some chemical facts:

  1. pH is a measure of acidity or alkalinity: the higher the pH the lower the acidity and the higher the alkalinity.
  2. The higher the concentration of hydroxide ions the lower the acidity or the higher the alkalinity of the solution, this is the higher the pH.

Hence, since you are asked to state the solution with the highest concentration of hydroxide ions, you just pick the highest pH. This is the option d, pH = 12.59.

These mathematical relations are used to find the exact concentrations of hydroxide ions:

  • pH + pOH = 14 ⇒ pOH = 14 - pH

  • pOH = - log [OH⁻] ⇒ [OH^-]=10^{-pOH}

Then, you can follow these calculations:

Solution    pH        pOH                            [OH⁻]

a.               3.21       14 - 3.21 = 10.79        antilogarithm of 10.79 = 1.6 × 10⁻¹¹

b.               7.00      14 - 7.00 = 7.00        antilogarithm of 7.00 = 10⁻⁷

c.                7.93      14 - 7.93 = 6.07        antilogarithm of 6.07 = 8.5 × 10⁻⁷

d.               12.59     14 - 12.59 = 1.41        antilogarithm of 1.41 = 0.039

e.               9.82      14 - 9.82 = 4.18        antilogarithm of 4.18 = 6.6 × 10⁻⁵

From which you see that the highest concentration of hydroxide ions is for pH = 12.59.

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