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Question: Would an aqueous solution of 0.25 M sodium formate be acidic, basic, or neutral? Write the equation for the salt hydrolysis and use it to justify your answer. Ka = 1.8 x 10^-4.-Free Course Hero Question Answer.

Question Description: Would an aqueous solution of 0.25 M sodium formate be acidic, basic, or neutral? Write the equation for the salt hydrolysis and use it to justify your answer. Ka = 1.8 x 10^-4.

Free Course Hero Answer

Course Hero Answer & Explanation:

Sodium formate (HCOONa) is salt of a weak acid ( formic acid, HCOOH) and strong base NaOH, so 0.25 M sodium formate solution will be basic.

The equation for salt hydrolysis is written as:

HCOO(aq) + H2O(aq) —> HCOOH (aq) + OH(aq)

pH  of 0.25 M sodium formate solution = 8.57

The pH of the solution is more than 7, so it is basic.Step-by-step explanation

Given:

The concentration of HCOONa = 0.25 M

Ka =1.8 x 10^-4

Sodium formate (HCOONa) is salt of a weak acid ( formic acid, HCOOH) and strong base NaOH, so 0.25 M sodium formate solution will be basic.

The equation for salt hydrolysis is written as:

HCOO(aq) + H2O(aq) —> HCOOH (aq) + OH(aq)

The concentration of HCOONa is given.  To know the aqueous solution of 0.25 M sodium formate is acidic, basic, or neutral, find out the pH of the solution. If the pH of the solution is 7, then it is neutral. If the pH of the solution is less than 7, then it is acidic and if the pH of the solution is more than 7, then it is basic.

Calculate Kb :

Kb = Kw/Ka 

Kb = (1.0 x10^-14)/( 1.8 x 10^-4) = 5.56 x10^-11

Sodium formate (HCOONa) is salt that dissociates into HCOO– and Naions and the pH of the solution is based on the hydrolysis of this salt. 

Construct an ICE table and calculate [OH]:

   HCOO(aq) + H2O(aq) —> HCOOH (aq) + OH(aq)

I   0.25                                         0                        0

C    -x                                       +x                       +x

E    (0.25 -x)                              x                         x

Kb = [HCOOH][OH]/ [HCOO]

5.56 x10^-11= x x/ (0.25 -x)

x^2 =5.56 x10^-11* 0.25

x^2  = 1.39 x 10^-11

x =3.73 x 10^-6  

x = [OH] = 3.73 x 10^-6 M

Calculate pOH:

pOH = – log [OH]

pOH = – log (3.73 x 10^-6) = 5.43

Calculate pH:

We know

pH + pOH =14.00

pH = 14.00 – pOH

pH = 14.00 – 5.43

pH = 8.57

The pH of the solution is more than 7, so it is basic.

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