Question

What is the pH of a solution formed when 42.0mL of 0.120 M NaOH is added to 50.0mL of 0.120 M CH3COOH? (Ka= 1.8x10^-5)

Answer 1

Calculations :-

(i) When 42.0 ml of 0.120 M NaOH is added to 50 ml of CH3COOH the salt 0.12 moles of CH3COONa is produced   as per reaction,

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

...............................1mole...................................................1.0 mole

therefore, 0.12mole.............................................0.12 mole ...

..................................................[ CH3COONa] = ( 0.12 / 92 ) x 1000

............................................................................= 1.3043M

(iv) Calculate the volume of acetic acid concentration added in excess = ( 50.0 - 42.0 ) = 8.0 ml.

....So , now 8.0 ml of 0.120 M CH3COOH is present in 92.0 ml of solution ( due to unreacted CH3COONa )

(v) Calculate concentration of acetic acid in this solution as,

...........................................................................[CH3COOH] =.( 8.00 x 0.92 ) /92

...................................................................................................= 0.08 M

(vi)

a. Now, the solution contains CH3COONa & CH3COOH ( ie . salt of a weak acid and strong base + the concerned weak acid ) therefore it should be regarded as a buffer solution whose pH is calculated using Handerson - Hasselbalch's equation ie.

............................pH = pKa + log [ salt ] / [ acid ]

b. calculate pH of this solution now by substituting the calculated values of

salt , [ CH3COONa] = 1.3043M , [CH3COOH] = 0.08M &

..........................pKa = -log Ka = { -log( 1.8 x 10^-5) }

....................................................= 4.7447

.....................................Thus,pH = 4.7447 + log ( 1.3043 / 0.08 )

.....................................................= 4.7447 + log(16.30375)

......................................................= 4.7447 + 1.21228

......................................................= 5.9569

...............................................or,~ = 5.96