What is the % dissociation of 0.01 N NH 4 OH in its buffer solution containing 0.01 N NH 4 Cl: ( K b . of .NH 4 OH =2 × 10-5)

Question

What is the % dissociation of 0.01 N NH 4 OH in its buffer solution containing 0.01 N NH 4 Cl: ( K b . of .NH 4 OH =2 × 10-5) (A) 0.2 % (B) 2 × 10-3 % (C) 2 × 10-5 (D) 5 × 10-3 %

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/395aa2bb0fa3d0f9ba564ec75fc366bc-.png" /> K b =((0.01 α)(0.01 α+0.01)/0.01(1-α)) 2 × 10-5=0.01 α(1+α) α=2 × 10-3 % α=2 × 10-3 × 100=0.2

Q. What is the $%$ dissociation of $0.01 N N H_{4} O H$ in its buffer solution containing $0.01 N N H_{4} Cl :$
$(K_{b}$ of $N H_{4} O H = 2 \times 1 0^{- 5})$

$0.2%$

$2 \times 1 0^{- 3} %$

$2 \times 1 0^{- 5}$

$5 \times 1 0^{- 3} %$

$K_{b} = \frac{( 0.01 α ) ( 0.01 α + 0.01 )}{0.01 ( 1 - α )}$
$2 \times 1 0^{- 5} = 0.01 α (1 + α)$
$α = 2 \times 1 0^{- 3}$
$% α = 2 \times 1 0^{- 3} \times 100 = 0.2$

Q. What is the % dissociation of 0.01NNH4​OH in its buffer solution containing 0.01NNH4​Cl: (Kb​ of NH4​OH=2×10−5)

0.2%

2×10−3%

2×10−5

5×10−3%