Question

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

20 mL of 0.001M HCl and 40 mL of 1.5M Acetic acid

20 mL of 0.001M HCl and 50 mL of 2.5M Sodium Acetate

Answer 1

HCl is strong acid compared to acetic acid and hence ionizes completely.

moles of HCl in 20ml of 0.001M= Molarity* Voliume (L) 0.001*20/1000=0.00002, moles of acetic acid = 1.5*40/1000 =0.06

Volume after mixing = 20+40= 60ml =60/1000 =0.06L

concentrations : HCl = 0.00002/0.06=0.00033, acetic acid =0.06/0.06=1

[H+] from HCl=0.00033, CH3COOH+ H2O----------->CH3COO- + H3O+, Ka= [CH3COO-][H3O+]/[CH3COOH]

let x= drop in concentration of Acetic acid to reach equilibrium

hence x²/(1-x)= Ka= 1.8*10^-5, when solved using excel, x= 4.25*10-3, [H3O+] = 4.25*10^-3 ,

total of H+ =0.00033 + 4.25*10^-3=0.00458, pH= -log(0.00458)= 2.34

2.

20 mL of 0.001M HCl and 50 mL of 2.5M Sodium Acetate

moles of HCl = 0.001*20/1000 = 0.00002, moles of sodium acetate= 2.5*50/1000 = 0.125

CH3COONa + HCl -------->CH3COOH + NaCl

limiting reactant is HCl. moles of acetic acid formed = 0.00002. moles of sodium acetate remaining =0.125-0.00002=0.12498

pH= pka+ log[ sodium acetate/ acetic acid ]

concentration ratio of sodium acetate/ acetic acid = mole ratio of sodium acetate/acetic acid

pH= 4.75+log (0.12498/0.00002)= 8.54