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Annette [7]
4 years ago
14

A 1.50 L buffer solution consists of 0.118 M butanoic acid and 0.318 M sodium butanoate. Calculate the pH of the solution follow

ing the addition of 0.063 moles of NaOH . Assume that any contribution of the NaOH to the volume of the solution is negligible. The K a of butanoic acid is 1.52 × 10 − 5 .
Chemistry
1 answer:
kaheart [24]4 years ago
5 0

Answer:

pH = 5.493

Explanation:

It is possible to find pH of a buffer using H-H equation:

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where [A⁻] is concentration of sodium butanoate and [HA] is concentration of butanoic acid.</em>

<em />

pKa is -log Ka. As Ka of butanoic acid is 1.52×10⁻⁵, <em>pKa is 4.818</em>

Before reaction, moles of butanoic acid and sodium butanoate are:

butanoic acid: 1.50L × (0.118mol / L) = <em>0.177 moles butanoic acid</em>

sodium butanoate: 1.50L × (0.318mol / L) = <em>0.477 moles sodium butanoate</em>

NaOH reacts with butanoic acid thus:

NaOH + butanoic acid → sodium butanoate + water.

<em>Butanoic acid decreases concentration and sodium butanoate increases concentration</em>

Thus, after reaction, moles of butanoic acid and sodium butanoate are:

butanoic acid: 0.177mol - 0.063mol = 0.114mol

sodium butanoate: 0.477mol + 0.063mol = 0.540mol

As total volume is 1.50L, concentrations are:

[A⁻] [sodium butanoate] = 0.540mol / 1.50L = <em>0.360M</em>

[HA] [butanoic acid] = 0.114mol / 1.50L = <em>0.076M</em>

<em />

Replacing in H-H equation:

pH = 4.818 + log₁₀ [0.360M] / [0.076M]

<em>pH = 5.493</em>

<em />

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Considering the definition of molar mass, the moles of gas used are 10.625 moles.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Amount of moles used</h3>

Natural gas has a molar mass of 16.0 g/mole.

You started out the day with a tank containing 200.0 g of natural gas.  So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 200 grams are contained in how many moles?

amount of moles at the beginning=\frac{200 gramsx1 mole}{16 grams}

<u><em>amount of moles at the beginning= 12.5 moles</em></u>

At the end of the day, your tank contains 30.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 30 grams are contained in how many moles?

amount of moles at the end=\frac{30 gramsx1 mole}{16 grams}

<u><em>amount of moles at the end= 1.875 moles</em></u>

The number of moles used will be the difference between the number of moles used initially and the contents at the end of the day.

moles used= amount of moles at the beginning - amount of moles at the end

moles used= 12.5 moles - 1.875 moles

<u><em>moles used= 10.625 moles</em></u>

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Finally, the moles of gas used are 10.625 moles.

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A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjug
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Answer:

ΔpH = 0.296

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Henderson-Hasselbalch formula to find pH in a buffer is:

pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

Replacing with known values:

5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

0.260 =  log₁₀ [CH₃COO⁻] / [CH₃COOH]

1.820 = [CH₃COO⁻] / [CH₃COOH] <em>(1)</em>

As total molarity of buffer is 0.100M:

[CH₃COO⁻] + [CH₃COOH] = 0.100M <em>(2)</em>

Replacing (2) in (1):

1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

2.820[CH₃COOH] = 0.100M

[CH₃COOH] = 0.100M / 2.820

[CH₃COOH] = <em>0.035M</em>

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5.40 mL of a 0.490 M HCl are:

0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = <em>7.429x10⁻³ moles of CH₃COO⁻</em>

<em />

Moles of CH₃COOH  before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

5.425x10⁻³ moles +  2.646x10⁻³ moles = <em>8.071x10⁻³ moles of CH₃COOH</em>. Replacing these values in Henderson-Hasselbalch formula:

pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

pH = 4.704

As initial pH was 5.000, change in pH is:

ΔpH = 5.000 - 4.740 = <em>0.296</em>

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3 years ago
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pashok25 [27]

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d1i1m1o1n [39]

Answer:

2.35 M

Explanation:

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