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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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<u>Answer:</u> The experimental van't Hoff factor is 1.21

<u>Explanation:</u>

The expression for the depression in freezing point is given as:

\Delta T_f=iK_f\times m

where,

i = van't Hoff factor = ?

\Delta T_f = depression in freezing point  = 0.225°C

K_f = Cryoscopic constant  = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21

Hence, the experimental van't Hoff factor is 1.21

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