12. Calculate the pH of a 0.10 M solution of potassium formate, KHCOO. Ka of formic...

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12. Calculate the pH of a 0.10 M solution of potassium formate,KHCOO. Ka of formic acid (HCOOH) = 1.8 x 10^-4. Write thedissociation equation and label each ion as acidic, basic, orneutral. Also write the hydrolysis equation.
Dissociation equation: _cation: anion:
Hydrolysis equation:_____________________________________

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Dissociation equation

Â Â HCOOH (aq) +Â Â H2O (l)Â Â Â --- > HCOO-(aq) + H3O+(aq)

IÂ Â Â 0.10Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

C -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +x

E (0.10-x)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x

Acidic species

H3O+]

Basic species

HCOO-

Ka expression

Ka = [HCOO-][H3O+] / [HCOOH]

1.8 x 10^-4 = x^2/ 0.10-x

x^2 = 1.8 E-4 * (010-x)

Lets solve this quadratic equation.

x = 0.004154

[H3O+]= 0.004154

pH = -log ([H3O+] = - log ( 0.004154)

= 2.38

pH of the solution would be 2.38

Cation is H3O+

anion is HCOO-

Hydrolysis equation of HCOO-

HCOO-(aq) + H2O (l) ---> HCOOH (aq) + OH-(aq)

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