An aqueous mixture at room temperature is 0.1 M with respect to ammonium chloride and 0.01 M with respect to NH4OH ; pKb of aqueous ammonia as a base is 5. The pH of the mixture is nearly

Question

An aqueous mixture at room temperature is 0.1 M with respect to ammonium chloride and 0.01 M with respect to NH4OH ; pKb of aqueous ammonia as a base is 5. The pH of the mixture is nearly (A) 7.5 (B) 6.8 (C) 6.5 (D) 8.0

Tardigrade Admin · Accepted Answer

pOH=pKb+log([. S a l t ]./[. B a s e ].) So 14-pH=5+log(0.1 M/0.01 M) =5+log10=5+1=6 or pH=14-6=8.0 .

Q. An aqueous mixture at room temperature is 0.1 M with respect to ammonium chloride and 0.01 M with respect to $N H_{4} O H$ ; $p K_{b}$ of aqueous ammonia as a base is 5. The pH of the mixture is nearly

7.5

6.8

6.5

8.0

$pO H = p K_{b} + l o g \frac{[ S a lt ]}{[ B a se ]}$

So $14 - p H = 5 + l o g \frac{0.1 M}{0.01 M}$

$= 5 + l o g 10 = 5 + 1 = 6$

or $p H = 14 - 6 = 8.0$ .

Q. An aqueous mixture at room temperature is 0.1 M with respect to ammonium chloride and 0.01 M with respect to NH4​OH ; pKb​ of aqueous ammonia as a base is 5. The pH of the mixture is nearly

7.5

6.8

6.5

8.0

pOH=pKb​+log[Base][Salt]​ So 14−pH=5+log0.01M0.1M​ =5+log10=5+1=6 or pH=14−6=8.0 .

Q. An aqueous mixture at room temperature is 0.1 M with respect to ammonium chloride and 0.01 M with respect to $N H_{4} O H$ ; $p K_{b}$ of aqueous ammonia as a base is 5. The pH of the mixture is nearly

$pO H = p K_{b} + l o g \frac{[ S a lt ]}{[ B a se ]}$

So $14 - p H = 5 + l o g \frac{0.1 M}{0.01 M}$

$= 5 + l o g 10 = 5 + 1 = 6$

or $p H = 14 - 6 = 8.0$ .