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3 years ago

Choose one:____.

Chemistry

1 answer:

mote1985 [20]3 years ago

5 0

Answer:

FeBr3 is the Lewis acid, and FeBr4-1 is the Lewis base.

Explanation:

A Lewis acid is any specie that has empty orbitals into which it can receive a lone pair of electrons.

A Lewis base is any specie that donates a lone pair of electrons to another to form a dative covalent bond.

If we look at the species listed in the options, FeBr3 is lone pair acceptor hence it is a Lewis acid while FeBr4- is a lone pair donor hence it is Lewis base.

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Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 × 10 − 5), with 0.1000 M NaOH

Zina [86]

Here is the full question

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 × 10 − 5), with 0.1000 M NaOH solution after the following additions of titrant (total volume of added base given):

a) 10.00 mL

pH =

b) 20.10 mL

pH =

c) 25.00 mL

pH =

Answer:

pH = 4.81

pH = 10.40

pH = 12.04

Explanation:

Number of moles of butanoic acid

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $10.00 \ mL * \frac{L}{1000 \ mL }* \frac{0.1000 \ mol }{L}$

= 0.001000 mol

pKa of butanoic acid = - log Ka

= - log ( 1.54 × 10⁻⁵)

= 4.81

Equation for the reaction is expressed as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

The ICE Table is expressed as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

Initial 0.002000 0.001000 0

Change - 0.001000 - 0.001000 + 0.001000

Equilibrium 0.001000 0 0.001000

Total Volume = (20.00 + 10.00 ) mL

= 30.00 mL = 0.03000 L

Concentration of [CH₃CH₂CH₂COOH] = $\frac{0.001000 \ mol}{ 0.03000 \ L }$

= 0.03333 M

Concentration of [CH₃CH₂COO⁻] = $\frac{0.001000 \ mol}{ 0.03000 \ L}$

= 0.03333 M

By Henderson- Hasselbalch equation

pH = pKa + log $\frac{conjugate \ base}{acid }$

pH = pKa + log $\frac{CH_3CH_2CH_2COO^-}{CH_3CH_2CH_2COOH}$

PH = 4.81 + log $\frac{0.03333}{0.03333}$

pH = 4.81

Thus; the pH of the resulted buffer solution after 10.00 mL of NaOH was added = 4.81

b )

After the equivalence point, we all know that the pH of the solution will now definitely be determined by the excess H⁺

Number of moles of butanoic acid

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $20.10 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002010 mol

Following the previous equation of reaction , The ICE Table for this process is as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

Initial 0.002000 0.002010 0

Change - 0.002000 -0.002000 + 0.002000

Equilibrium 0 0.000010 0.002000

We can see here that the base is present in excess;

NOW, number of moles of base present in excess

= ( 0.002010 - 0.002000) mol

= 0.000010 mol

Total Volume = (20.00 + 20.10 ) mL

= 40.10 mL × $\frac{1 \ L}{1000 \ mL }$

= 0.04010 L

Concentration of acid [OH⁻] = $\frac{0.000010 \ mol}{0.04010 \ L }$

= $2.494*10^{-4} M$

Using the ionic product of water:

$[H_3O^+] = \frac{K \omega }{[OH^-]}$

where

$K \omega = 10^{-14}$

$[H_3O^+] = \frac{1.0*10^{-14}}{2.494*10^{-14}}$

= $4.0*10^{-11}M$

pH = - log $[H_3O^+}]$

pH = - log [ $4.0*10^{-11}M$ ]

pH = 10.40

Thus, the pH of the solution after the equivalence point = 10.40

After the equivalence point, pH of the solution is determined by the excess H⁺.

Number of moles of butanoic acid

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $25.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002500 mol

From our chemical equation; The ICE Table can be illustrated as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

Initial 0.002000 0.002500 0

Change - 0.002000 -0.002000 +0.002000

Equilibrium 0 0.000500 0.002000

Base is present in excess

Number of moles of base present in excess = [ 0.002500 - 0.002000] mol

= 0.000500 mol

Total Volume = ( 20.00 + 25.00 ) mL

= 45.00 mL

= 45.00 × $\frac{1 \ L}{1000 \ mL }$

= 0.04500 L

Concentration of acid [OH⁻] = $\frac{0.0005000 \ mol}{ 0.04500 \ L }$

= 0.01111 M

Using the ionic product of water $[H_3O^+] = \frac{K \omega }{[OH^+]}$

= $\frac{1.0*10^{-14}}{0.01111}$

= $9.0*10^{-13}$ M

pH = - log $[H_3O^+}]$

pH = - log [ $9.0*10^{-13}M$ ]

pH = 12.04

Thus, the pH of the solution after the equivalence point = 12.04

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Why would the white blood cells have an abundance of lysosomes?

77julia77 [94]

Answer:

Explanation:

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Lysosomes are found in all animal cells, but are most numerous in disease-fighting cells, such as white blood cells. This is because white blood cells must digest more material than most other types of cells in their quest to battle bacteria, viruses, and other foreign intruders.

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3 years ago

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Arte-miy333 [17]

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In an eluent solution of 70% hexanes and 30% ethyl acetate, compound A moves farther on the TLC plate than compound B. Compound

8090 [49]

Answer:

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Explanation:

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RSB [31]

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