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pH of a solution of 0.1 M [CH3COONH4(aq)...

pH of a solution of 0.1 M `[CH_3COONH_4(aq)]` is `[given: K_a(CH_3COOH) = K_b(NH_4OH) = 1.8 x 10^-5)]`

A

1

B

9

C

7

D

13

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% hydrolysis of 0.1M CH_(3)COONH_(4), when K_(a)(CH_(3)COOH)=K_(b)(NH_(4)OH)=1.8xx10^(-5) is:

Both conc. Being 1 M, the factor by which the percent hydrolysis of CH_3COO^- is greater in CH_3COONH_4 than in CH_3COONa is : [K_a(CH_3COOH)=K_b(NH_4OH)=1.8xx10^(-5)]

Calculate the pH at the equivalence point when a solution of 0.1 M CH_3COOH is titrated with a solution of 0.1 M NaOH. K_a(CH_3COOH)= 1.8 xx 10^(-5)

The pH of a solution containing 0.1 M CH_3COONa and 0.1 M (C_2H_5COO)_2 Ba will be K_a(CH_3COOH)=2xx10^(-5). K_a(C_2H_5COOH)=1.5xx10^(-5) :

Calculate the pH of a solution containing 0.1 M CH_(3)COOH and 0.15 M CH_(3)COO^(-) . (K_(a) "of" CH_(3)COOH=1.8xx10^(-5))

Carefully observe the given figure and using data provided find the EMF of shown Galvenic cell in volt: Solution A is 0.1M each in NH_(4)OH and NH_(4)CI and solution B is 0.1M each in CH_(3)COOH and CH_(3)COONa^(+) . [Given: K_(a)(CH_(3)COOH) = 10^(-5), K_(b) (NH_(4)OH) = 10^(-5) and (2.303RT)/(F) = 0.06 volt]

What is [H^(+)] in mol//L of a solution that is 0.20 M in CH_(3)COONa and 0.1 M in CH_(3)COOH ? K_(a) for CH_(3)COOH is 1.8xx10^(-5) ?

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