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NEET NEET Questions — Free MCQ Practice

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If pKb for fluoride ion at $ 25 ^\circ C $ is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is

A. $ 1.74 \times 10 ^ {-3} $ B. $ 3.52 \times 10 ^ {-3} $ C. $ 6.75 \times 10 ^ {-4} $ D. $ 5.38 \times 10 ^ {-2} $

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Explanation

The ionization constant of hydrofluoric acid (HF) can be found using the relationship between the ionization constant of the acid (Ka) and the base dissociation constant (Kb) of its conjugate base (fluoride ion, F^-). The relationship is given by: $$ K_a imes K_b = K_w $$ where $$ K_w $$ is the ion-product constant for water at 25Â°C ($1 imes 10^{-14}$). Given $$ pK_b = 10.83 $$, we first find $$ K_b $$: $$ K_b = 10^{-pK_b} = 10^{-10.83} = 1.48 imes 10^{-11} $$ Then, $$ K_a = rac{K_w}{K_b} = rac{1 imes 10^{-14}}{1.48 imes 10^{-11}} = 6.75 imes 10^{-4} $$ Hence, the correct answer is $ 6.75 imes 10 ^ {-4} $.

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The PH of a 0.01 M solution of acetic acid having degree of dissociation 1.25% is

A. 5.623 B. 2.903 C. 3.723 D. 4.503

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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

A.

$ 4 \times 10 ^ {-8} $

B.

$ 8 \times 10 ^ {-12} $

C.

$ 12 \times 10 ^ {-12} $

D.

$ 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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