Calculate the pH of a buffer which is 0.1 M in acetic acid and 0.15 M in sodium acetate. Given that the ionization constants of acetic acid is 1.75xx10^(-5). Also calculate the change in pH of the buffer if to 1 litre of the buffer (i) 1 cc of 1 M NaOH are added (ii) 1 cc of 1 M HCl are added. Assume that the change in volume is negligible. What will be the buffer index of the above buffer ?

Calculate the pH of a buffer which is 0.1 M in acetic acid and 0.15 M

1.	Calculate the pH of a buffer which is 0.1 M in acetic acid and 0.15 M in sodium acetate. Given that the ionization constants of acetic acid is 1.75xx10^(-5). Also calculate the change in pH of the buffer if to 1 litre of the buffer (i) 1 cc of 1 M NaOH are added (ii) 1 cc of 1 M HCl are added. Assume that the change in volume is negligible. What will be the buffer index of the above buffer ?
Answer» Solution :`pH = pK_(a) + log. (["Salt"])/(["Acid"]) = - log (1.75 xx 10^(-5)) + log. (0.15)/(0.10)` `=(5-0.2430) + 0.1761=4.757+0.1761=4.933`. (i) 1 CC of 1 M NaOH contains NaOH `= 10^(-13)` MOL. This will convert `10^(-3)` mol of acetic acid into the salt so that salt formed `=10^(-3)` mol Now,[Acid]=0.10 - 0.001 = 0.099 M [Salt]=0.15+0.001 = 0.51 M `pH = 4.757 + log. (0.151)/(0.099) = 4.757 + 0.183 = 4.940` `:.` Increase in pH = 4.933 = 0.007 which is negligible. (ii) 1 cc of 1 M HCl contains HCl = `10^(-3)` mol. This will convert `10^(-3)` mol `CH_(3)CO ONa ` into `CH_(3)CO OH`. `:. ` Now, [Acid]=0.10 + 0.001 = 0.101 M [Salt] = 0.15 - 0.001 = 0.19 M `:. pH = 4.757 + log. (0.149)/(0.101) = 4.757+0.169= 4.926` `:. Decrease in pH = 4.933 - 4.926 = 0.007which is again negligible. (III) Calculation of buffer INDEX ltbr. No. of moles of HCl or NaOH added = 0.001 mol Change in pH = 0.007 Hence, buffer index `=(dn)/(dpH)=(0.001)/(0.007 ) = (1)/(7) = 0.143`