Question

You titrate 10.0 mL of 0.30 M acetic acid with 0.10 M sodium hydroxide.

a. What is the pH of the solution after you have added 10.00 mL of NaOH?
b. What is the pH of the solution at the equivalence point?

Answer 1

1)

Calculate number of moles of CH3COOH in 10 ml. of 0.3M CH3COOH as

= [ (10 x 0.3) /1000 ] = 0.00 3 moles

Similarly moles of NaOH in 10 ml of 0. 10 NaOH = [ ( 10 x 0.1 ) / 1000 ]

...............................................................................= 0.001 moles

Study the stoichiometry of neutralization reaction occuring during titration, ie.

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

No. of moles ....................................1.......................1.............. ..................1

So, equal number of moles of NaOH and acetic acid react to form an equal number of

moles of CH3COONa

We have 0.001 moles of NaOH ( which is the limiting reagent in this case ) ,producing 0.001moles of CH3COONa

Again the total volume of solution after adding 10 ml of o.1 M NaOH becomes = 20. 0ml

therefore concentration of CH3COONa in solution after adding 10 ml of NaOH, ie.[CH3COONa] = ( 0.001 x 1000) / 20

........................= 0.05 M

& the moles of CH3COOH present in 20ml of solution at this stage

= moles of CH3COOH added - moles of CH3COOH used for reaction = ( 0.003 - 0.001)

....................................................................................................... .........= 0.002 moles

hence , [CH3COOH ] ={( 0.002 x 1000 ) / 20 } = 0.10 M

The solution now contains CH3COONa ( [ CH3COONa ] = 0.05M) & CH3COOH with
[CH3COOH ] = 0.1M so apply Handerson's Hasselbalch equation to calculate pH of

the solution -

pH = pKa + log {[ CH3COONa] / [CH3COOH]}

......= 4.7447 + log (0.05 / 0.10 )............(.pKa of CH3COOH is Known to be = 4.7447)

...... = 4.4436

or, ~ = 4.44

2)

At equivalence point the volume of 0.1M required = (0.3 x 10)/ 0.1 = 30 ml

and the total volume of reaction mixture / solution = 30.0 +10.0 = 40.0 ml

therefore, moles of NaOH present in 40ml of solution =( 0.1 x40 ) /1000

..................................................................................... = 0.004 moles

As per stoichiometry 0.004moles of NaOH will produce 0.004 moles of CH3COONa

so [ CH3COONa ] in solution =( 0.004 x 1000 ) / 40

.............................................. = 0.10M

The solution at equivalence point , now , will contain no H⁺ ,except those yielded due to

hydrolysis of the salt CH3COONa. Hence pH of the solution is calculated using equation-

pH = 7 + 1/2 pKa + 1/2 log C

...... = 7 + 1/2 (4.7447 ) + 1/2 log ( 0.10)

...... = 8.8873

...or, ~= 8.89