Match the following - If ka of HCN - 5 - 10-10- ka of HOCN - 3-2 - 10-4 - kb of NH3 - 1-8 - 10-5 - ka of CH3COOH - 1-8 - 10-5 all data at 250 C

(A)anionic hydrolysis leads to formation of Oh due to which of ph gt; 7 #160;(B) (N_1V_1)_NaOH gt; (N_2 V_2 )_HCl (C) strong acid + weak acid , so p ...

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Byju's Answer

Standard XII

Chemistry

Salt of Strong Acid and Strong Bases

Match the fol...

Question

Match the following :
If $k_{a}$ of HCN $= 5 \times 10^{- 10}$ , $k_{a}$ of HOCN $= 3.2 \times 10^{- 4}, k_{b} o f N H_{3} = 1.8 \times 10^{- 5}, k_{a} o f C H_{3} C O O H = 1.8 \times 10^{- 5}$ ( all data at $25^{0} C)$

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Solution

(A)anionic hydrolysis leads to formation of Oh due to which of ph > 7
(B) $(N_{1} V_{1})_{N a O H} > (N_{2} V_{2})_{H C l}$
(C) strong acid + weak acid , so pH < 7
(D) hydrolysis of cation leads to the formation of $H^{+}$ due to which pH < 7

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Similar questions

Q. The correct descending order of the heat liberated (in

k J

) during the neutralisation of the acids

C H_{3} C O O H (W), H F (X), H C O O H (Y)

and

H C N (Z)

under identical conditions is:

(

K_{a}

C H_{3} C O O H = 1.8 \times 10^{- 5}, H C O O H = 1.8 \times 10^{- 4}, H C N = 4.9 \times 10^{- 10}

and

H F = 3.2 \times 10^{- 4}

)

K_{a}

for

H C N

and

{C H}_{3} C O O H

are

4.9 \times 10^{- 10}

and

1.8 \times 10^{- 5}

respectively. Calculate the equilibrium constant for the reaction:

{C H}_{3} C O O H + N a C N ⇌ {C H}_{3} C O O N a + H C N

Q. 19. What is [CN-] in a solution that contain 0.02 M HCN (Ka = 1.8 X 10 exp.(-5) and 0.01 M HCL ? 1. 8 X 10 exp(-6) 2. 1.8 X 10 exp(-5) 3. 9 X 10 exp(-5) 4. 3.6 X 10 exp (-5)

Q. A mixture of weak acid is 01M in

H C O O H

(K_{a} = 1.8 \times 10^{- 4})

and 0.1M in

H O C N

(K_{a} = 3.3 \times 10^{- 4})

. Hence,

[H_{3} O^{+}]

is:

Which of the following indicates the correct order of acidic strength of the following acids.

I. HCOOH (Ka = 1.8 $\times$ $10^{- 4}$ )

II. $C H_{3} C O O H$ (Ka = 1.8 $\times$ $10^{- 5}$ )

III : HCN (Ka = 4.8 $\times$ $10^{- 10}$ )

IV: HS- (Ka = 1 $\times$ 10-14)

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Match the following : If ka of HCN =5×10−10, ka of HOCN =3.2×10−4,kbofNH3=1.8×10−5,kaofCH3COOH=1.8×10−5( all data at 250C)

(A)anionic hydrolysis leads to formation of Oh due to which of ph > 7 (B) (N1V1)NaOH>(N2V2)HCl(C) strong acid + weak acid , so pH < 7(D) hydrolysis of cation leads to the formation of H+ due to which pH < 7

Match the following :
If $k_{a}$ of HCN $= 5 \times 10^{- 10}$ , $k_{a}$ of HOCN $= 3.2 \times 10^{- 4}, k_{b} o f N H_{3} = 1.8 \times 10^{- 5}, k_{a} o f C H_{3} C O O H = 1.8 \times 10^{- 5}$ ( all data at $25^{0} C)$

(A)anionic hydrolysis leads to formation of Oh due to which of ph > 7
(B) $(N_{1} V_{1})_{N a O H} > (N_{2} V_{2})_{H C l}$
(C) strong acid + weak acid , so pH < 7
(D) hydrolysis of cation leads to the formation of $H^{+}$ due to which pH < 7