The pH of a buffer solution is 4.745. When 0.044 mole of Ba(OH)₂ is added to 1 lit. of the buffer, the pH changes to 4.756. Then the buffer capacity is

0.044 mol of Ba(OH)₂ = 0.044 x 2 mol of OH^-

When 0.088 mol of OH^- is added, change in P^H

                                                 = (4.756 - 4.745)

                                                 = 0.011 

Conversely,

0.011 change of P^H occurs when moles of OH^- added = 0.088 mol

1 change of P^H occurs when moles of OH^- added $=\frac{0.088}{0.011}\times 1=8\mathrm{mol}$