Question

Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or with a bicarbonate ion (pka of h2co3 = 6.37)?

Answer 1

Phenol, pentane-2,4-dione and diethyl malonate are strong enough to react completely with a hydroxide ion. But none of these compounds can react with a bicarbonate ion.

Further Explanation:

The formula to calculate ${\text{p}}{{\text{K}}_{\text{a}}}$ of acid is as follows:

${\text{p}}{{\text{K}}_{\text{a}}} = - \log {{\text{K}}_{\text{a}}}$       …… (1)

                                                       

Here,

${{\text{K}}_{\text{a}}}$ is the dissociation constant of acid.

Equation (1) implies more the value of ${{\text{K}}_{\text{a}}}$, smaller will be the ${\text{p}}{{\text{K}}_{\text{a}}}$ of the acid and vice-versa. The higher value of ${{\text{K}}_{\text{a}}}$ means the acid is completely dissociated into its respective ions when dissolved in the solvent and vice-versa. The compounds that have ${\text{p}}{{\text{K}}_{\text{a}}}$ values less than that of the given solvent will dissociate in it and thereby reacting with it and vice-versa.

The value of ${\text{p}}{{\text{K}}_{\text{a}}}$ of water is 15.74. The value of ${\text{p}}{{\text{K}}_{\text{a}}}$ of phenol is 9.89. The value of ${\text{p}}{{\text{K}}_{\text{a}}}$ of pentane-2,4-dione is 9. The value of ${\text{p}}{{\text{K}}_{\text{a}}}$ of diethyl malonate is 13.

All these compounds have ${\text{p}}{{\text{K}}_{\text{a}}}$ values less than that of water so these can react with hydroxide ion completely. But the value of ${\text{p}}{{\text{K}}_{\text{a}}}$ of ${{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is 6.37. All the given compounds have ${\text{p}}{{\text{K}}_{\text{a}}}$ values greater than that of ${{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ so these cannot react with a bicarbonate ion.

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Answer details:

Grade: High School

Chapter: Acid, base and salts

Subject: Chemistry

Keywords: Ka, pKa, acids, phenol, pentane-2,4-dione, diethyl malonate, 13, 15.74, 6.37, 9, 9.89, H2CO3, bicarbonate ion, water, value, hydroxide ion, compounds.

Answer 2

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

Further explanation

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

$K~=~10^{-pK}$

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

• 1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

$K~=~10^{4.9}$

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

$K`=~10^{1.12]$

Reaction can take place

• 2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

$K~=~10^{ -3.56}$

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

$K`=~10 ^{-12.93}$

Reaction does not happen

• 3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

$K~=~10^{-3.26}$

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

$K~=~ 10^{-12.63}$

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

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Keywords : acid base reaction, the equilibrium constant