Test: Equllibrium (September 18) - NEET MCQ

Given s = 0.5 × 10^–4 moles/lit

but pOH+ pH = 14  or pOH = 14 – pH

14 – 9.25 – 0 = pK_b pK_b = 4.75

For reaction to proceed from right to left

i.e the reaction will be fast in backward direction i.e r_b > r_f.

The higher is the tendency to donate proton, stronger is the acid. Thus the correct order is R – COOH > HOH > R – OH > CH ≡ CH depending upon the rate of donation of proton.

K_sp for AgI = 1 × 10^–16 In solution of KI, I^– would be due to the both AgI and KI, 10^–4 solution KI would provide = 10^–4 I^–
AgI would provide, say = x I^– (x is solubility of AgI)

as x is very small ∴ x² can be ignored

∴      10^–4 x = 10^–16

For 

For an acid-base indicator

Taking negative on both sides

or we can write  

The solution formed from isomolar solutions of sodium oxide, sodium sulphide, sodium selenide H₂O, H₂S, H₂Se & H₂Te respectively.
As the acidic strengh increases from H₂O to H₂Te thus pH decreases and hence the correct of pHs is pH₁ > pH₂ > pH₃ > pH₄.

First option is incorrect as the value of K_P given is wrong. It should have been

For a solution of 10^–8 M HCl [H⁺] = 10^–8 [H⁺] of water = 10^–7
Total [H⁺] = 10^–7 + 10^–8 = 10 × 10^–8 + 10^–8
10^–8 (10 + 1) = 11 × 10^–8

HNO₂ is a weak acid and NaNO₂ is salt of that weak acid and strong base (NaOH).

Given,

We have to calculate

For this equation,

Now operate,

[H₃O]⁺ for a solution having pH = 3 is given by [H₃O]⁺ = 1×10^–3 moles/litre  

Similarly for solution having pH = 4, [H₃O]⁺ = 1 × 10^–4 moles/ litre and for pH=5 [H₃O⁺] = 1×10^–5 moles/ litre Let the volume of each solution in mixture be IL, then total volume of mixture solution  L = (1 + 1 + 1) L =3L
Total [H₃O]⁺ ion present in mixture solution = (10^–3 + 10^–4 + 10^–5) moles Then [H₃O]⁺ ion concentration of mixture solution

=  0.00037 M = 3.7 ×10^–4 M.

The highest pH will be recorded by the most basic solution. The basic nature of hydroxides of alkaline earth metals increase as we move from Mg to Ba and thus the solution of BaCl₂ in water will be most basic and so it will have  highest pH.

Given reaction are

Let the total pressure for reaction (i) and (ii) be P₁ and P₂ respectively, then

After dissociation,  

At equilibrium (1–α) α α
[Let 1 mole of X dissociate with α as degree of dissociation ]
Total number of moles =   1– α + α + α = (1+α)

Thus  

We have,

Dividing (i) by (ii), we get

i.e. Option (c) is correct answer.