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melomori [17]
3 years ago
7

An analytical chemist is titrating 118.3 mL of a 0.3500 M solution of butanoic acid (HC3H7CO2) with a 0.400 M solution of KOH. T

he pKa of butanoic acid is 4.82. Calculate the pH of the acid solution after the chemist has added 115.4 mL of the solution to it. Note for advanced students: you may assume the final volume equals the initial volume of KOH the solution plus the volume of solution added. Round your answer to 2 decimal places.
Chemistry
1 answer:
TEA [102]3 years ago
4 0

Answer:

pH = 12.33

Explanation:

Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.

The titration reaction is

HA + KOH ---------------------------- A⁻ + H₂O + K⁺

number of moles of HA :   118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA

number of  moles of OH  : 115.4 mL/1000ml/L x 0.400 mol/L  = 0.046 mol A⁻

therefore the weak acid will be completely consumed and what we have is  the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.

n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH

pOH = - log (KOH)

M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M

pOH = - log (0.0021) = 1.66

pH = 14 - 1.96 = 12.33

Note: It is a mistake to ask for the pH of the <u>acid solutio</u>n since as the above calculation shows we have a basic solution the moment all the acid has been consumed.

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If I mix bleach and namona and throw it on a person what will happen​
satela [25.4K]

Answer:

They have the chance to inhale toxic fumes secreted by the mixture.

Explanation:

7 0
3 years ago
The free energy change for the following reaction at 25 °C, when [Cr3+] = 1.32×10-3 M and [Fe3+] = 1.14 M, is 131 kJ: Cr3+(1.32×
larisa [96]

Answer:

E°cell = - 1.3575 V

This reaction is spontaneous in the reverse direction

Explanation:

The given cell reaction:

Cr³⁺(1.32 × 10⁻³ M) + Fe²⁺(aq) → Cr²⁺(aq) + Fe³⁺(1.14 M)

The given Gibbs free energy: ΔG = 131 kJ = 131 × 10³ J     (∵ 1 kJ = 10³ J)

As we know,

ΔG = - n F E°cell

Here, n - the number of moles of electrons transferred = 1

F - Faraday constant = 96500

E°cell - cell potential = ?

\therefore E^{\circ }_{cell} = -\frac{\Delta G}{n \: F} = -\frac{131\times 10^{3}\, J}{1\, mol\times96500 \, C.mol^{-1}}

\Rightarrow E^{\circ }_{cell} = -1.3575\, V

<u>For a given chemical reaction if-</u>

1. ΔG = negative and E°cell = positive

⇒ <em>The reaction is spontaneous and proceeds spontaneously in the forward direction.</em>

2.  ΔG = positive and E°cell = negative

⇒ <em>The reaction is non-spontaneous and proceeds spontaneously in the reverse direction.</em>

<u>Since, for this chemical reaction: </u>

Cr³⁺(1.32 × 10⁻³ M) + Fe²⁺(aq) → Cr²⁺(aq) + Fe³⁺(1.14 M)

ΔG = + 131 × 10³ J ⇒ positive

and, E°cell = - 1.3575 V ⇒ negative

<u>Therefore, this reaction is spontaneous in the reverse direction.</u>

6 0
3 years ago
Which of the following is not an example of a pure element?
motikmotik
Copper wire is not an example of a pure element because although it's made by pure elements, it's not one itself. It's made by factories.

Have a nice day! :)
8 0
3 years ago
Read 2 more answers
Calculate the adjusted retention times for both pentafluorobenzene and benzene if the elution time for unretained solute is 1.06
vovangra [49]

Answer:

Pentafluorobenzene: 11,92 min

Benzene: 12,14 min

Explanation:

<em>Retention time of pentafluorobenzene is 12,98 min and 13,20 min of benzene.</em>

The adjusted retention time is the time an analyte spends in the column not the stationary phase. As time of unretained solute is 1,06 min the adjusted retention time for an analyte is:

tr' = tr - 1,06min

For pentafluorobenzene:

tr' = 12,98min - 1,06min = <em>11,92 min</em>

For benzene:

tr' = 13,20 - 1,06min = <em>12,14 min</em>

<em></em>

I hope it helps!

3 0
3 years ago
How many grams of sodium are in 0.820 moles of na2so4?
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<span>the molar mass of a compound is the sum of the products of the atomic masses by the number of atoms of the element.
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1 mol of </span>Na₂SO₄<span> has a mass of 142 g.
In 1 mol of </span>Na₂SO₄<span> the mass of Na is 23 g/mol x 2 = 46 g.
                             
Mass of Na in 1 mol of </span>Na₂SO₄ is - 46 g
                           
mass of Na in 0.820 mol of Na₂SO₄ - 46 g /1 mol x 0.820 mol = 37.72 g.
mass of Na is 37.72 g
5 0
3 years ago
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