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Alenkasestr [34]
3 years ago
13

Why carbon monoxide is not acidic ​

Chemistry
1 answer:
mojhsa [17]3 years ago
7 0

Answer:

Well, carbon monoxide can be created from formic acid by adding sulphuric acid which will dehydrate said formic acid:

HCOOH

−

→

−

−

−

H

2

SO

4

CO+H

2

O

HCOOH→HX2SOX4CO+HX2O

Therefore, we can imagine the reverse reaction theoretically, which would make carbon monoxide an acidic oxide. However, the forward reaction does not proceed easily and it needs both the high acidity of sulphuric acid and its strong dehydrative properties to actually work. And your question mentions using hot, concentrated sodium hydroxide to make the reverse one work.

Most oxides that are classified as acidic or basic either have a very electrophilic central atom (e.g.  

CO

2

COX2

) which can be attacked by the weak nucleophile water (which in turn can then release an acidic proton), or they have a high charge density on the oxygen which allows it to abstract a proton from water directly. Carbon monoxide is neither. If you check out its molecular orbitals, you will notice that even though carbon is partially positive it has the largest HOMO contribution, meaning a proton would be more likely to attatch to the carbon side — which doesn’t want one at all. The LUMO is, luckily, also more carbon-centred, meaning nucleophilic attacks on carbon are possible. However, it is also degenerate due to the double bond so that an attack is not favoured.

Thus, the carbon monoxide molecule is one that won’t react with water at all and totally defies the concept of acidic/basic oxides.

Abbreviations:

HOMO is a widely used abbreviation for the Highest Occupied Molecular Orbital, i.e. the one with the highest energy that still contains electrons. It is usually the orbital that will attack nucleophilicly or that will be attacked electrophilicly.

LUMO is a widely used abbreviation for the Lowest Unoccupied Molecular Orbital, i.e. the virtual (unoccupied) orbital that has the lowest energy. When considering a nucleophilic attack, the attacking electrons will usually interact with the LUMO. Electrophiles attack with other molecules’ HOMO with their LUMO.

Explanation:

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4 years ago
A buffer solution contains 0.496 M hydrocyanic acid and 0.399 M sodium cyanide . If 0.0461 moles of sodium hydroxide are added t
pochemuha

Answer : The pH of the solution is, 9.63

Explanation : Given,

The dissociation constant for HCN = pK_a=9.31

First we have to calculate the moles of HCN and NaCN.

\text{Moles of HCN}=\text{Concentration of HCN}\times \text{Volume of solution}=0.496M\times 0.225L=0.1116mole

and,

\text{Moles of NaCN}=\text{Concentration of NaCN}\times \text{Volume of solution}=0.399M\times 0.225L=0.08978mole

The balanced chemical reaction is:

                          HCN+NaOH\rightarrow NaCN+H_2O

Initial moles     0.1116       0.0461     0.08978

At eqm.       (0.1116-0.0461)    0       (0.08978+0.0461)

                        0.0655                       0.1359

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

Now put all the given values in this expression, we get:

pH=9.31+\log (\frac{0.1359}{0.0655})

pH=9.63

Therefore, the pH of the solution is, 9.63

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7 0
3 years ago
Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the adde
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Answer: The molar mass of the unknown compound is 200 g/mol

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=2.74^0C = Depression in freezing point

i= vant hoff factor = 1 (for molecular compound)

K_f = freezing point constant = 5.12^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (benzene)= 0.250 kg  

Molar mass of solute = M g/mol

Mass of solute  = 26.7 g

2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}

M=200g/mol

Thus the molar mass of the unknown compound is 200 g/mol

4 0
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