Calculate PH Of 0.01 M acetic acid . `K_a = 1.8 xx 10^-5` at 298 K.

For a 0.01 (N) solution of sodium acetate, calculate pH. Given for acetic acid K_a = 1.9 xx 10^(-5) .

Determine the degree of hydrolysis and PH of 0.02 M of sodium acetate. (Given k_a = 1.8 xx 10^-5 , K_w = 1 xx 10^-14 )

What is the pH of a 1 M CH_3COONa solution? K_2 of acetic acid = 1.8 xx 10^-5 and K_w = 10^-14 mol^2 litre^-2:

Conductivity of 0.00241 M acetic acid solution is 7.896 xx 10^(-5) S cm^(-1) . Calculate its molar conductivity in this solution. If Lambda_(m)^(@) for acetic acid be 390.5 S cm^(2) "mol"^(-1) , what would be its dissociation constant? (i) First , find molar conductivity using the formula,, Lambda_(m) = ( K xx 1000)/( C ) (ii) Then find degree of dissociation ( alpha) and dissociation constant ( K_(a)) by using formula alpha = ( Lambda_(m))/( Lambda_(m)^(@)) and K_(a) = ( C alpha^(2))/( 1-alpha) respectively.

Conductivity of 0.00241M acetic acid is 7.896 xx 10^(-5) S cm^(-1) . Calculate its molar condcutivity. If Lambda_(m)^(@) acetic acid is 390.5 S cm^(2) "mol"^(-1) , what is its dissociation constant?

The pH of simple sodium acetate and acetic acid buffer is given by, pH= pK_a + log [Salt]/[Acid] K_a of acetic acid = 1.8 xx 10^-5. If [Salt] = [Acid] =0.1 M, the pH of the solution would be about:

What is the PH OF 0.1 M HCl.

The equivalent conductance of 0.1 N acetic acid is 5 cm^2 ohm^-1 gm eq^-1 and at infinite dilution is 390 cm^2ohm^-1gm eq^-1 . Calculate the degree of dissociation of acetic acid.

PH of 0.01 M HCl solution is ______.