Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Chemical Equilibrium question

2021 · 25 Jul · Shift 2 · Q21
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Chemical Equilibrium
  5. /2021 · 25 Jul · Shift 2 · Q21

Chemical Equilibrium question

2021 · 25 Jul · Shift 2 · Q21

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
Assuming that Ba(OH)2Ba(OH)_2Ba(OH)2​ is completely ionised in aqueous solution under the given conditions the concentration of H3O+H_3O^+H3​O+ ions in 0.005 M aqueous solution of Ba(OH)2Ba(OH)_2Ba(OH)2​ at 298 K is ‾×\underline{\hspace{2cm}}\times​× 10 −-− 12 mol L −-− 1. (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 1

  1. Write the dissociation of Ba(OH)2Ba(OH)_2Ba(OH)2​

Since Ba(OH)2Ba(OH)_2Ba(OH)2​ is assumed to be completely ionised:

Ba(OH)2→Ba2++2OH−Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-Ba(OH)2​→Ba2++2OH−

  1. Calculate [OH−][OH^-][OH−]

Given concentration of Ba(OH)2Ba(OH)_2Ba(OH)2​ is 0.005 M0.005\,M0.005M. Each mole of Ba(OH)2Ba(OH)_2Ba(OH)2​ gives 222 moles of OH−OH^-OH−, so

[OH−]=2×0.005=0.010 M=10−2 M[OH^-] = 2 \times 0.005 = 0.010\,M = 10^{-2}\,M[OH−]=2×0.005=0.010M=10−2M

  1. Use ionic product of water at 298 K298\,K298K

At 298 K298\,K298K,

Kw=[H3O+][OH−]=10−14K_w = [H_3O^+][OH^-] = 10^{-14}Kw​=[H3​O+][OH−]=10−14

Therefore,

[H3O+]=10−1410−2=10−12 M[H_3O^+] = \frac{10^{-14}}{10^{-2}} = 10^{-12}\,M[H3​O+]=10−210−14​=10−12M

  1. Match with the asked format

The question asks:

[H3O+]=‾×10−12 mol L−1[H_3O^+] = \underline{\hspace{1cm}} \times 10^{-12}\,\text{mol L}^{-1}[H3​O+]=​×10−12mol L−1

Since

[H3O+]=1×10−12 mol L−1[H_3O^+] = 1 \times 10^{-12}\,\text{mol L}^{-1}[H3​O+]=1×10−12mol L−1

the required integer is:

1\boxed{1}1​

PreviousNext

More from Chemical Equilibrium

  • The OH − concentration in a mixture of 5.0 mL of 0.0504 M NH4​Cl and 2 mL of 0.0210 M NH3​ solution is x × 10 − 6 M. The value of x is ​. (Nearest integer) [Given Kw = 1 × 10 − 14 and Kb =…2021 · Numerical
  • The equilibrium constant Kc at 298 K for the reaction A + B ⇌ C + D is 100. Starting with an equimolar solution with concentrations of A, B, C and D all equal to 1M, the equilibrium concentration of D is ​…2021 · Numerical
  • The reaction rate for the reaction [PtCl4​]2−+ H2​O ⇌[Pt(H2​O)Cl3​]− + Cl− was measured as a function of concentrations of different species. It was observed that dt−d[[PtCl4​]2−]​=4.8×10−5[[PtCl4​]2−]−2.4×10−3[[Pt(H2​O)Cl3​]−][Cl−]…2021 · Numerical
  • A homogeneous ideal gaseous reaction AB2(g)​⇌A(g)​+2B(g)​ is carried out in a 25 litre flask at 27 ∘ C. The initial amount of AB2​ was 1 mole and the equilibrium pressure was 1.9 atm. The value…2021 · Numerical
  • The number of moles of NH3​, that must be added to 2L of 0.80 M AgNO3​ in order to reduce the concentration of Ag+ ions to 5.0 × 10 − 8 M (Kformation for [Ag(NH3​)2​]+ = 1.0 × 108) is ​.…2021 · Numerical
  • When 5.1 g of solid NH4​HS is introduced into a two litre evacuated flask at 27 ∘ C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The Kp for the reaction at 27 ∘ C is x × 10 − 2. The…2021 · Numerical
  • PCl5​ ⇌ PCl3​ + Cl2​ Kc = 1.844 3.0 moles of PCl5​ is introduced in a 1 L closed reaction vessel at 380 K. The number of moles of PCl5​ at equilibrium is ​× 10 − 3. (Round…2021 · Numerical
  • The equilibrium constant for the reaction A(s) ⇌ M(s) +21​ O2​(g) is Kp = 4. At equilibrium, the partial pressure of O2​ is ​ atm. (Round off to the nearest integer)2021 · Numerical