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wariber [46]
2 years ago
12

Describe how you would prepare exactly 100 mL of 0.109 M picolinate buffer, pH 5.61. Possible starting materials are pure picoli

nic acid (pyridine-2-carboxylic acid, FM 123.10), 1.0 M HCl, and 1.0 M NaOH. Place the given steps in order. Not all of the steps will be used.
Chemistry
1 answer:
Pepsi [2]2 years ago
3 0

Answer:

1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL

Explanation:

<em>The pKa of the picolinic acid is 5.4.</em>

Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:

pH = pKa + log [Picolinate] / [Picolinic]

<em>Where [] could be taken as moles of each species</em>

<em />

5.61 = 5.4 + log [Picolinate] / [Picolinic]

0.21 = log [Picolinate] / [Picolinic]

1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>

<em></em>

Now, both picolinate and picolinic acid will be:

0.100L * (0.109mol / L) =

0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>

<em></em>

First, as we will start with picolinic acid, we need add:

0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid

Now, replacing (2) in (1):

1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]

1.62181 [Picolinic] = 0.0109 moles - [Picolinic]

2.62181 [Picolinic] = 0.0109 moles

[Picolinic] = 4.157x10⁻³ moles

And:

[Picolinate] = 0.0109 - 4.157x10⁻³ moles =

<h3>6.743x10⁻³ moles</h3><h3 />

To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:

Picolinic acid + NaOH → Picolinate + Water

<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>

<em />

6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:

6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =

<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />

Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:

<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>
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<h3>Further explanation</h3>

Given

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Required

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There are 3 parts in this question:
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2) To find the final Boyle's constant k_{f}
3) To verify whether gas is obeying Boyle's law or not

Given data:
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The initial pressure(in atmospheric pressure) = P_{i} = 4.0 atm

The final pressure(in atmospheric pressure) = P_{f} = 8.0 atm
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First you need to know what Boyle's law is:
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</span>
The Mathematical form of Boyle's law is:
P =  \frac{k}{V}

Where,
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V = Volume of the gas
k = Boyle's constant

Now let's solve aforementioned parts one by one:
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The initial pressure(in atmospheric pressure) = P_{i} = 4.0 atm
The Boyle's constant = k_{i} = ?

According to the Boyle's law,

P_{i} = \frac{k_{i}}{V_{i}}

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2.
The final pressure(in atmospheric pressure) = P_{f} = 8.0 atm
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According to the Boyle's law,

P_{f} = \frac{k_{f}}{V_{f}}

=> k_{f} =  P_{f}V_{f}
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Ans-2) k_{f} = 48

3.
In order to verify Boyle's law, the initial Boyle's constant should be EQUAL to the final Boyle's constant, meaning:

k_{i} = k_{f}

Since,
k_{i} = 48
k_{f} = 48

Therefore,
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Ans-3) Hence proved: The gas IS obeying the Boyle's law.

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