Question

A 27 mL solution of .100 M CH3COOH is titrated with a .220 M KOH solution. Calculate the pH after the following additions of the KOH solution:

a) 0 mL

b) 5 mL

Answer 1

a)

[OH-] from KOH = 0

                            CH3COOH < ---> CH3COO- + H+

initial:                   0.1    

at equilibrium: 0.1 -x                             x               x

From this table : http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

Ka of CH3COOH = 1.76*10^-5

Ka= [CH3COO-][H+]/[CH3COOH]

1.76*10^-5 = x*x /(0.1-x)

sincxe Ka is small 0.1 - x will tend to 0.1

1.76*10^-5 = x*x /(0.1

x=1.33*10^-3 M =[H+]

pH= -log [H+] = -log (1.33*10^-3 ) = 2.88

b)

• The number of millimoles of CH3COOH to be neutralized is 27?0.1=2.7 mmol CH3COOH
• The number of millimoles of OH^_ that will be added within 5 mL is 5?.0.22=1.1 mmol OH?

                            CH3COOH +OH- < ---> CH3COO- + H2O

initial:                   2.7 1.1    0 0

at equilibrium: 2.7-1.1=1.6         0              1.1              -

However, this only gives us the millimoles. To get the concentration we must divide by the total volume. The total volume is the 27 mL original soultion of CH3COOH plus the 5 mL of KOH that was added. Therefore, the total volume is 27mL+5mL=32mL

• Concentration of CH3COOH = 1.6mmol/ 32mL=0.05 M                 
• Concentration of CH3COO- = 1.2mmoL / 32mL=0.0375M

pKa of CH3COOH =4.75

USE:

pH=pka+log( [A?]/[HA] )

pH = 4.75 + log (0.0375/0.05)

     =4.63