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Chemical Equilibrium question

2021 · 26 Aug · Shift 1 · Q16
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  5. /2021 · 26 Aug · Shift 1 · Q16

Chemical Equilibrium question

2021 · 26 Aug · Shift 1 · Q16

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
The OHOHOH −-− concentration in a mixture of 5.0 mL of 0.0504 M NH4ClNH_4ClNH4​Cl and 2 mL of 0.0210 M NH3NH_3NH3​ solution is x ×\times× 10 −-− 6 M. The value of x is ‾\underline{\hspace{2cm}}​. (Nearest integer) [Given Kw = 1 ×\times× 10 −-− 14 and Kb = 1.8 ×\times× 10 −-− 5]
Numerical answer
View written solutionFree

Correct answer: 3

  1. Identify the system

We have a mixture of a weak base NH3NH_3NH3​ and its salt NH4ClNH_4ClNH4​Cl. So this is a basic buffer:

NH3+H2O⇌NH4++OH−NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-NH3​+H2​O⇌NH4+​+OH−

Given:

  • Volume of NH4Cl=5.0 mLNH_4Cl = 5.0\,\text{mL}NH4​Cl=5.0mL, concentration =0.0504 M= 0.0504\,M=0.0504M
  • Volume of NH3=2.0 mLNH_3 = 2.0\,\text{mL}NH3​=2.0mL, concentration =0.0210 M= 0.0210\,M=0.0210M
  • Kb(NH3)=1.8×10−5K_b(NH_3)=1.8\times 10^{-5}Kb​(NH3​)=1.8×10−5
  • Kw=1.0×10−14K_w=1.0\times 10^{-14}Kw​=1.0×10−14

We need [OH−][OH^-][OH−] in the final mixture.

  1. Calculate moles of base and conjugate acid

For NH4ClNH_4ClNH4​Cl:

n(NH4+)=0.0504×5.0×10−3n(NH_4^+) = 0.0504 \times 5.0\times 10^{-3}n(NH4+​)=0.0504×5.0×10−3 =2.52×10−4 mol= 2.52\times 10^{-4}\,\text{mol}=2.52×10−4mol

For NH3NH_3NH3​:

n(NH3)=0.0210×2.0×10−3n(NH_3)=0.0210\times 2.0\times 10^{-3}n(NH3​)=0.0210×2.0×10−3 =4.20×10−5 mol=4.20\times 10^{-5}\,\text{mol}=4.20×10−5mol

  1. Use the buffer relation for a basic buffer

For the equilibrium

NH3+H2O⇌NH4++OH−NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-NH3​+H2​O⇌NH4+​+OH−

Kb=[NH4+][OH−][NH3]K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}Kb​=[NH3​][NH4+​][OH−]​

Hence,

[OH−]=Kb⋅[NH3][NH4+][OH^-]=K_b\cdot \frac{[NH_3]}{[NH_4^+]}[OH−]=Kb​⋅[NH4+​][NH3​]​

Since both are in the same final volume, the ratio of concentrations equals the ratio of moles:

[OH−]=Kb⋅n(NH3)n(NH4+)[OH^-]=K_b\cdot \frac{n(NH_3)}{n(NH_4^+)}[OH−]=Kb​⋅n(NH4+​)n(NH3​)​

Substitute values:

[OH−]=1.8×10−5×4.20×10−52.52×10−4[OH^-]=1.8\times 10^{-5}\times \frac{4.20\times 10^{-5}}{2.52\times 10^{-4}}[OH−]=1.8×10−5×2.52×10−44.20×10−5​

Now,

4.20×10−52.52×10−4=4.2025.2=16=0.1667\frac{4.20\times 10^{-5}}{2.52\times 10^{-4}}=\frac{4.20}{25.2}=\frac{1}{6}=0.16672.52×10−44.20×10−5​=25.24.20​=61​=0.1667

Therefore,

[OH−]=1.8×10−5×0.1667[OH^-]=1.8\times 10^{-5}\times 0.1667[OH−]=1.8×10−5×0.1667 =3.0×10−6 M=3.0\times 10^{-6}\,M=3.0×10−6M

So,

[OH−]=x×10−6 M[OH^-]=x\times 10^{-6}\,M[OH−]=x×10−6M

Hence,

x=3x=3x=3

  1. Comparison with stored correct answer

Stored correct answer = 333

Our derived answer = 333

So the answer agrees.

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