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Equilibrium MCQs for NEET — Chemistry Questions with Answers

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By adding 20 ml 0.1 N HCl to 20 ml 0.1 N KOH the $ P^H $ of obtained solution will be :

A. 0 B. 7 C. 2 D. 9

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Explanation

When equal volumes and concentrations of a strong acid (HCl) and a strong base (KOH) are mixed, they neutralize each other completely. The reaction is: $$ HCl + KOH ightarrow KCl + H_2O $$ Since they neutralize each other, the resulting solution is neutral, meaning the pH will be 7 at 25Â°C. Hence, the correct answer is 7.

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If the $ K_b $ value in the hydrolysis reaction $ B^+ + H_2 O \rightleftharpoons BOH + H^+ is 1.0 \times 10 ^ {-6} $ then hydrolysis constant of salt would be

A. $ 1.0 \times 10 ^ {-6} $ B. $ 1.0 \times 10 ^ {-7} $ C. $ 1.0 \times 10 ^ {-8} $ D. $ 1.0 \times 10 ^ {-9} $

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For a sparingly soluble salt $ A_pB_q relation ship of its solubility product(K_{SP})$ with its solubility(s) is.

A. $ K_{SP} = S ^ {p+q} P ^ p q ^ q $ B. $ K_{SP} = S ^ {p+q} P ^ q q ^ p $ C. $ K_{SP} = S ^ {pq} P ^ p q ^ q $ D. $ K_{SP} = S ^ {pq} ( pq) ^ {p+q} $

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Explanation

$ ApBq_{(s)} \rightleftharpoons PA^{q+}_(aq) + q B ^{p-} _{(aq)} $ Solubility is PS qS mol/lit $K_{SP} = (PS) ^ P \times (qs) ^ q$ $ = S^ (p+q) \times P^P \times q^ q $

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How many grams $ CaC_2O_4 (mw = 128) on dissolving in distill water will give saturated solution [K_{SP} = (CaC_2O_4 ) = 2.5 \times 10 ^ {-9} mol^2 l^{-2 } $

A. 0.0064 gm B. 0.1280 gm C. 0.01268 gm D. 1.2800 gm

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Explanation

To find the amount of CaC_2O_4 that will dissolve to form a saturated solution, we use the solubility product constant ( extit{K}{sp}). For CaC_2O_4, extit{K}{sp} = 2.5 imes 10^{-9}. The solubility (s) can be found from extit{K}_{sp} = s^2, giving s = rac{1}{2} imes 10^{-4.5}. The molar mass of CaC_2O_4 is 128 g/mol. Therefore, the mass of CaC_2O_4 that will dissolve in 1 liter of water is s imes 128 = 0.0064 g.

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If the concentration of $ CrO_4 { 2–} ion ina saturated solution of silver chromate is 2 \times 10 ^ {-4} $ solubility product of silve chromate will be

$ 4 \times 10 ^ {-8} $

$ 8 \times 10 ^ {-12} $

$ 12 \times 10 ^ {-12} $

$ 32 \times 10 ^ {-12} $

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Explanation

The equilibrium is:
$A g_{2} C r O_{4} â‡Œ 2 A g^{+} + C r O_{4 2-}$

On the basis of this equation, the concentration of

C r O_{4 2-}

ion will be half of the concentration of

A g^{+}

ions.

Thus,

[A g^{+}] = 4 \times 10^{-4} M

and

[C r O_{4 2-}] = 2 \times 10^{-4} M

K_{s p} = [A g^{+}]^{2} [C r O_{4 2-}] = 32 \times 10^{-12}

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According to bronsted- lowryconcept. correct order of relative strength of bases follows the order

A. $ Cl^- \gt CH_3 COO^- \gt OH ^ - $ B. $ Cl ^ - \gt OH^- \gt CH_3 COO ^ - $ C. $ OH^ - \gt CH_3 COO \gt Cl ^ - $ D. $ OH ^ - \gt Cl ^ - \gt CH_3 COO ^ - $

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Explanation

According to the Bronsted-Lowry concept, base strength is determined by the ability to accept protons. The hydroxide ion (OH^-) is a stronger base than acetate (CH_3COO^-) and chloride (Cl^-). Thus, the correct order of base strength is OH^- > CH_3COO^- > Cl^-.

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$ HSO_4 ^ - + OH ^ - \rightarrow SO^{2-} _ 4 + H_2O $ Which is correct about conjugate acid base pair

A. $ HSO ^ - _4 is Conjugate acid of base SO^{2-} _4 $ B. $ HSO_4 ^ - is Conjugate base of acid SO^{2-} _ 4 $ C. $ SO_4 ^ {2-} is Conjugate base of acid HSO ^- _4 $ D. None of these

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Explanation

In the given reaction, $HSO_4^- ightarrow SO_4^{2-} + H_2O$, $HSO_4^-$ acts as an acid as it donates a proton (H^+), and $SO_4^{2-}$ is its conjugate base. Thus, $SO_4^{2-}$ is the conjugate base of the acid $HSO_4^-$. This is because in a conjugate acid-base pair, the acid has one more proton (H^+) than its conjugate base.

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Which of following base is weakest

A. $ NH_4OH:K_b 1.6 \times 10 ^{-6} $ B. $ C_6H_5 NH_2 : K_b = 3.8 \times 10 \times 10 ^ {-10} $ C. $ C_2 H_5 NH_2 : K_b = 5.6 \times 10 ^ {-4} $ D. $ C_2 H_7 N : K_b = 6.3 \times 10 ^{-10} $

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Explanation

The strength of a base is determined by its $K_b$ value. A lower $K_b$ value indicates a weaker base. Given the $K_b$ values:

$NH_4OH: K_b = 1.6 imes 10^{-6}$
$C_6H_5NH_2: K_b = 3.8 imes 10^{-10}$
$C_2H_5NH_2: K_b = 5.6 imes 10^{-4}$
$C_2H_7N: K_b = 6.3 imes 10^{-10}$ The weakest base is $C_6H_5NH_2$ because it has the smallest $K_b$ value of $3.8 imes 10^{-10}$.

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HClO is a weak acid. concentration of $ H^+ ion in 0.1 M solution of HClO (Ka = 5 \times 10^{-8} ) $ will be

A. $ 7.07 \times 10 ^ {-5} M $ B. $ 5 \times 10 ^ {-9} M $ C. $ 5 \times 10 ^ {-7} M $ D. $ 7 \times 10 ^ {-4} M $

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Explanation

$ Co = 0.1 M Ka = 5 \times 10 ^{-8} $ $ [H ^ + ]= \sqrt {Ka.CO} $ $ = (5 \times 10 ^ {-8} \times 0.1) ^{ 1/2} = (50 \times 10 ^ {-10} ) ^ {1/2} = 7.07 \times 10 ^ {-5 } M. $

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$ NH_4Cl$ is acidic because

On hydrolysis $ NH_4Cl $ give weak base $ NH_4OH $ and strong acid HCl

Nitrogen donates a pair of $ e^- $

It is a salt of weak acid and strong base

On hyrdolysis $ NH_4Cl $ gives strong base and weak acid

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Explanation

$NH_4Cl$ is acidic because, upon hydrolysis, it gives a weak base ($NH_4OH$) and a strong acid (HCl). The hydrolysis reaction is: $NH_4Cl + H_2O ightarrow NH_4OH + HCl$ Since $HCl$ is a strong acid and completely dissociates in water, it makes the solution acidic.

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