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Delvig [45]
3 years ago
11

The following equation shows the equilibrium in an aqueous solution of ammonia: NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)NH3(aq)+H2O(l)⇌NH

4+(aq)+OH−(aq) Which of the following represents a conjugate acid-base pair?a) NH3 and H2O b) NH4+ and OH− c) H2O and OH− d) NH3 and OH−
Chemistry
1 answer:
yarga [219]3 years ago
8 0

Answer:

c) H2O and OH−

Explanation:

Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.

When an acid donates a proton, it changes into a base which is known as its conjugate base.

Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.

The acid and the base which is only differ by absence or presence of the proton are known as acid conjugate base pair.

Thus,  for the reaction,

NH_3+H_2O\rightleftharpoons Nh_4^++OH^-

The base is NH_3 and the conjugate acid of the base is NH_4^+.

Also, The base is OH^- and the conjugate acid of the base is H_2O.

<u>Correct option is :- c) H2O and OH−</u>

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1. 100gm of a 55% (M/M) nitric acid solution is to be diluted to 20% (M/M) nitric acid.
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Answer:

The volume of water to be added is 0.175 liters of water

Explanation:

The given concentration of the nitric acid = 55% (M/M)

The mass of the nitric acid solution = 100 gm

The concentration solution is to diluted to = 20% (M/M)

The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution

Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get

Let "x" represent the volume of the resulting solution, we have;

20% of x = 55 g of nitric acid

∴ 20/100 × x = 55 g

x = 55 g × 100/20 =  275 g

The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid

The mass of extra water to be added = 275 g - 100 g = 175 g

Volume = Mass/Density

The density of water ≈ 1 g/ml

∴ The volume of water to be added that gives 175 g of water =  175 g/(1 g/ml) = 175 ml. = 0.175 l

The volume of water to be added = 0.175 liters of water.

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Empirical formula mass

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