Calculate (H+) and % dissociation of 0.1 M solution of NH4OH solution. The ionisation constant for NH4OH is, Kb=2.0× 10- 5

Question

Calculate (H+) and % dissociation of 0.1 M solution of NH4OH solution. The ionisation constant for NH4OH is, Kb=2.0× 10- 5 (A) 7.09× 10- 12M,3 % (B) 7.09× 10- 12M,1.4 % (C) 9.02× 10- 12M,2.4 % (D) 9.02× 10- 12M,3 %

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-vvsux1uafbbknycs.jpg" /> K b =(( C α/ C (1-α))=( C α2/1-α) negligible (1 - α )=1 Kb=Cα 2 α =√(kb/C)=√(2 × 1 0- 5/0.1)=√2 × 1 0- 4 α =1.41× 10- 2 α =1.4 % [O H-]=Cα =0.1× 1.4× 10- 2 =1.4× 10- 3M [H-]=(Kw/[O H-])=(1 0- 14/1.4 × 1 0- 3) =7.09× 10- 12M

Q. Calculate $(H^{+})$ and % dissociation of 0.1 M solution of $N H_{4} O H$ solution. The ionisation constant for $N H_{4} O H$ is, $K_{b} = 2.0 \times 1 0^{- 5}$

$7.09 \times 1 0^{- 12} M, 3%$

$7.09 \times 1 0^{- 12} M, 1.4%$

$9.02 \times 1 0^{- 12} M, 2.4%$

$9.02 \times 1 0^{- 12} M, 3%$

Q. Calculate (H+) and % dissociation of 0.1 M solution of NH4​OH solution. The ionisation constant for NH4​OH is, Kb​=2.0×10−5

7.09×10−12M,3%

7.09×10−12M,1.4%

9.02×10−12M,2.4%

9.02×10−12M,3%

Kb​=C(1−α)(Cα​=1−αCα2​ negligible (1−α)=1 Kb​=Cα2 α=Ckb​​​=0.12×10−5​​=2×10−4​ α=1.41×10−2 α=1.4% [OH−]=Cα=0.1×1.4×10−2 =1.4×10−3M [H−]=[OH−]Kw​​=1.4×10−310−14​ =7.09×10−12M

Q. Calculate $(H^{+})$ and % dissociation of 0.1 M solution of $N H_{4} O H$ solution. The ionisation constant for $N H_{4} O H$ is, $K_{b} = 2.0 \times 1 0^{- 5}$

$7.09 \times 1 0^{- 12} M, 3%$

$7.09 \times 1 0^{- 12} M, 1.4%$

$9.02 \times 1 0^{- 12} M, 2.4%$

$9.02 \times 1 0^{- 12} M, 3%$