Question

QUESTION 10 Calculate the change in pH if 10mL of 0.1M HCl is added to the buffer made by mixing 25 mL of 1.0 M CH3COOH and 25 mL of 0.5 M CH3COONa. For these types of questions where a strong acid or base is added to a buffer solution, first write out the chemical reaction. What will the HCl react with: CH3COOH or CH3COO−? Since HCl is an acid, it will react with CH3COO-. Because HCl is a strong acid, it will completely react with CH3COO- to form CH3COOH. One approach to these problems is to make an ICE table and to work in mols. Complete the table below. Once the table is complete use the equation for [H3O+] in question 15. Q16 ICE table.png

Answer 1

pH =Pka + log (salt/acid)

sodium acetate is a salt of weak acid and strong base

CH3COOH is weak acid

moles of sodium acetate= 0.5*25/1000 =0.0125 moles

moles of acetic acid = 1*25/1000 =0.025

Volume of mixture= 25+25 =50 ml

concentration of acetic acid in the mixture = 0.025*1000/50=0.5 M

Concentration of sodium acetate =0.0125*1000/50= 0.25

ka of acetic acid = 1.8*10-5

pKa= -log (1.8*10-5)=4.744

pH= 4.744 +log (0.25/0.5)=4.74-0.30= 4.44

when 10 ml of 0.1 HCl is added, moles of HCl consmed= 0.1*10/1000 =0.001 moles

the following reaction takes place CH3COONa+HCl----> CH3COOH+ NaCl

CH3COONa gets consumed and CH3COOH gets added. Limting reactant is HCl since is it is only 0.001 moles while CH3COONa= 0.0125 moles

CH3COONa remaining =0.0125-0.001 = 0.0115

CH3COOH formed= 0.025+0.001 =0.026

concentration after mixting

CH3COONa =0.0115/60

CH3COOH= 0.026/60

pH= 4.74+log (0.015/0.026)=4.74-0.23889= 4.50

Chane in pH= 4.5-4.4 =0.06