Question

Calculate the pH of the solution when the following substances are added together ​

20 mL of 0.1M NaOH and 8 mL of 0.25 M Sodium Acetate

20 mL of 0.1 M NaOH and 8 mL of 0.25 M Acetic acid

Answer 1

sodium hydroxide is strong base and sodium acetate is weak base.

moles of NaOH in 20ml of 0.1M= 0.1*20/1000 =0.002 and sodium acetate= 0.25*8/1000 = 0.002

Volume after mixing = 20+8= 28ml= 28/1000= 0.028L

Concentrations : NaOH= 0.002/0.028=0.0714

NaOH ionizes completely because it is a strong base. NaOH----> Na+ + OH-

[OH-]= 0.714, pOH= 1.15, pH= 14-1.15= 12.85

2. CH3COOH + NaOH-------->CH3COONa + H2O

moles of CH3COONa formed= 0.002

concentration = 0.002/0.028=0.0714M

CH3COONa+ H2O ------->CH3COO- + OH-

Kb= [CH3COO-] [OH-]/[CH3COONa]

let x= drop in concentration of CH3COONa to reach equilibirum

at equilibirum [CH3COO-] =[OH-] =x and [CH3COONa]= 0.0714-x

Ka of acetic acid =1.8*10-5, Kb=10^-14/(1.8*10^-5)= 5.55*10^-10

x²/(0.0714-x)= 5.55*10^-10, when solved using excel, x= 6.3*10^-6, pOH= 5,2, pH= 15-5.2= 8.8