Question

calculate the amount of (NH4)2SO4 which must be added to 500ml of 0.2 (M) NH3 to yield a solution of pH = 9.35 , pKb(NH4)OH = 4.74 Please explain briefly

Answer 1

pH = 9.35 => pOH = 14-9.35 = 4.65.

We know, pOH = pKb - log(base/salt)

Therefore, log(base/salt) = pKb-pOH = 4.74-4.65 = 0.09.

So base/salt = 1.23.

Number of moles of NH3 = 0.2 * 0.5 = 0.1 mole. So, 0.1/ salt = 1.23, Concentration of Salt. = 0.08.

Hence, 0.08 mole of ammonium ion = 0.04 mole [NH4]2SO4.