1. When adding strong base to a buffer, why can we assume the equilibrium value of the OH- is zero?

When adding strong base to a buffer, we assume the equilibrium value of the hydroxyl ion (OH-) is zero because all hydroxyl ion (OH) react with hydrogen ion (H+) forming water. So there is no hydroxyl ion (OH-) left and values of hydroxyl ion (OH-) is become zero.

Hydrogen ion is released by weak acid which is present in the buffer solution.

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in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm