Question

what is the pH of the solution obtained by mixing 50.00 mL of 0.1250 M HOAc and 25.00 mL of 0.100 M NaOH?

Answer 1

Moles H⁺ = 50.0mL x 0.125M = 6.25 mmol

Moles OH^- = 25.00mL x 0.100M = 2.5 mmol

	HAc +	OH- =>	Ac-	+ H2O
I	6.25mmol	2.5mmol	------
C	- 2.5mmol	- 2.5mmol	+2.5mmol
E	0	0	+2.5 mmol

Now the acetate ion will reestablish the equilibrium as CH3COOH cant dissociate completly in to its constituent ion. The equilibrium is as follows:

Ac-

+ H2O

<==> HAc

+ OH-

Now, Find the concentration of the acetate ion using the total volume.

(add the ml of acid and mL of the base)

Molarity =	mol	=	mmol
	liters		mL

[Ac-] =	2.5mmol
	50.0ml + 25.0ml

[Ac^-] = 0.03 M

  

Ac-

+ H2O

<==> HAc

+ OH-

I 0.03 M .....        

C - X + X + X

E 0.03 - X + X + X

Write the equilibrium expression

Kb = { [HAc] [OH-] } / [ Ac-] - (1)

Solve for Kb , Kb = Kw / Ka , Kb = 10^-14 / 1.8 x 10^{-5 =} 5.56 x 10^-10

Plug values into the Kb expression (1)

5.56 x 10^-10 = [X] [X] / [0.03 - X]

X = [OH-] = 4.08 x 10^-6 M

Solve for pOH

pOH = -log[OH^-]= -log 4.08 x 10^-6 M , pOH = 5.38

Solve for pH

pH = 14 - 5.38 = 8.62