Question

calculate the ph of the following solution: 15.00ml of 0.1400 M Na Oh has been added to 25.00 00ml of a buffer which is 0.2500 M HAc and .2000 M NaAc. Ka HAc= 1.8x10^-5

Answer 1

Initial moles of acid = 0.2500 * 25.00 / 1000 = 0.006250 mol

Initial moles of salt = 0.2000 * 25.00 / 1000 = 0.00500 mol

Moles of Base added = 0.1400 * 15.00 / 1000 = 0.002100 mol

When a strong base is added to a buffer solution, It abstracts proton of equilibrium of weak acid, to form salt. Hence concentration of salt will increase and acid will decrease.

Final moles of acid = 0.006250 - 0.00200 = 0.004250 mol

final moles of salt = 0.00500 - 0.00200 = 0.00300 mol

Final [acid] = 0.004250 * 25.00 / (40.00) = 0.002656 M

Final [salt] = 0.00300 * 25.00 / 40.00 = 0.001875 M

And , pKa = - Log Ka = - Log(1.8 * 10-5) = 4.745

Formula,

pH = pKa + Log[salt]/[acid]

pH = 4.745 + Log(0.001875 / 0.002656)

pH = 4.594