3 g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20 mL of this solution `(1)/(2)` mL of 5 M NaOH is added . The pH of the solution is __________.
[Given : pKa a acetic acid = 4.75, molar mass of acetic acid = 60 g/mol, log 3 = 0.4771]
Neglect any changes in volume.

Answer

5.23

Explanation

mmoles of `CH_(3)COOH=50`
mmoles of HCl=25
mmoles of `CH_(3)COOH` in `20ml= 50/500 xx20=2`
mmoles of HCl in 20mL`=25/500 xx20=1`
Now `NaOH + HCl to NaCl + H_(2)O`
t=0 2.5 1
`t_(end)` 1.5 0
` CH_(3)COOH + NaOH to CH_(3)COONa + H_(2)O`
t=0 2 1.5 0
`t_(end)` 0.5 0 1.5
In the end we have an acidic buffer
`pH = pK_(a) + "log"(["salt’])/(["Acid’])`
`pH=5.23`