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Chemical Equilibrium question

2021 · 27 Aug · Shift 1 · Q21
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  5. /2021 · 27 Aug · Shift 1 · Q21

Chemical Equilibrium question

2021 · 27 Aug · Shift 1 · Q21

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
The number of moles of NH3NH_3NH3​, that must be added to 2LLL of 0.80 M AgNO3AgNO_3AgNO3​ in order to reduce the concentration of Ag+Ag^+Ag+ ions to 5.0 ×\times× 10 −-− 8 M (Kformation for [Ag(NH3)2]+[Ag(NH_3)_2]^+[Ag(NH3​)2​]+ = 1.0 ×\times× 108) is ‾\underline{\hspace{2cm}}​. (Nearest integer) [Assume no volume change on adding NH3NH_3NH3​]
Numerical answer
View written solutionFree

Correct answer: 4

  1. Write the complex formation equilibrium
Ag++2NH3⇌[Ag(NH3)2]+Ag^+ + 2NH_3 \rightleftharpoons [Ag(NH_3)_2]^+Ag++2NH3​⇌[Ag(NH3​)2​]+

Given formation constant,

Kf=[Ag(NH3)2+][Ag+][NH3]2=1.0×108K_f=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.0\times 10^8Kf​=[Ag+][NH3​]2[Ag(NH3​)2+​]​=1.0×108
  1. Initial moles of AgNO3AgNO_3AgNO3​

Volume =2 L=2\,L=2L, molarity =0.80 M=0.80\,M=0.80M

initial moles of Ag+=0.80×2=1.6 mol\text{initial moles of } Ag^+ = 0.80\times 2 = 1.6\text{ mol}initial moles of Ag+=0.80×2=1.6 mol

Since volume remains 2 L2\,L2L, total silver concentration is

[Ag]total=0.80 M[Ag]_{\text{total}}=0.80\,M[Ag]total​=0.80M
  1. Free Ag+Ag^+Ag+ concentration required
[Ag+]=5.0×10−8 M[Ag^+] = 5.0\times 10^{-8}\,M[Ag+]=5.0×10−8M

So almost all silver is present as complex. Hence,

[Ag(NH3)2+]=0.80−5.0×10−8≈0.80 M[Ag(NH_3)_2^+] = 0.80 - 5.0\times 10^{-8} \approx 0.80\,M[Ag(NH3​)2+​]=0.80−5.0×10−8≈0.80M
  1. Find equilibrium concentration of free NH3NH_3NH3​

Using

Kf=[Ag(NH3)2+][Ag+][NH3]2K_f=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}Kf​=[Ag+][NH3​]2[Ag(NH3​)2+​]​ 108=0.80(5.0×10−8)[NH3]210^8 = \frac{0.80}{(5.0\times 10^{-8})[NH_3]^2}108=(5.0×10−8)[NH3​]20.80​

Rearrange:

[NH3]2=0.80108×5.0×10−8[NH_3]^2 = \frac{0.80}{10^8\times 5.0\times 10^{-8}}[NH3​]2=108×5.0×10−80.80​ [NH3]2=0.805=0.16[NH_3]^2 = \frac{0.80}{5}=0.16[NH3​]2=50.80​=0.16 [NH3]=0.40 M[NH_3]=0.40\,M[NH3​]=0.40M
  1. Calculate total NH3NH_3NH3​ needed

Each mole of complex consumes 2 moles of NH3NH_3NH3​.

Complex concentration is about 0.80 M0.80\,M0.80M, so bound ammonia concentration is

2×0.80=1.60 M2\times 0.80=1.60\,M2×0.80=1.60M

Also, free ammonia concentration required at equilibrium is 0.40 M0.40\,M0.40M.

Thus total ammonia concentration to be added:

[NH3]added=1.60+0.40=2.00 M[NH_3]_{\text{added}} = 1.60+0.40=2.00\,M[NH3​]added​=1.60+0.40=2.00M

For 2 L2\,L2L solution,

moles of NH3=2.00×2=4.0 mol\text{moles of } NH_3 = 2.00\times 2 = 4.0\text{ mol}moles of NH3​=2.00×2=4.0 mol
  1. Final answer

Nearest integer:

4\boxed{4}4​
  1. Comparison with stored answer

Stored correct answer = 4, which matches our result.

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