How many grams of dipotassium phthalate (242.3 g/mol) must be added to 100 mL of 0.125 M potassium hydrogen phthalate to give a buffer of pH 5.80? The Ka's for phthalic acid are 1.12 x 10-3 and 3.90 x 10-6.
Ka2 = 3.9*10^-6
PKa2 = -logKa2
= -log(3.9*10^-6)
= 5.41
no of moles of KHP = molarity * volume in L
= 0.125*0.1 = 0.0125moles
dipotassium phthalate (K2P)
H2P(aq) -----------> H^+ + HP^- Ka1
HP^- ----------> H^+ + P^2- Ka2
PH = PKa + log[P^2-]/[HP^-]
5.80 = 5.41 + log[P^2-]/0.0125
log[P^2-]/0.0125 = 5.80-5.41
log[P^2-]/0.0125 = 0.39
[P^2-]/0.0125 = 2.4547
[P^2-] = 2.4547*0.0125
[P^2-] = 0.0307 moles
no of moels of K2P( dipotassium phthalate) = 0.0307moles
mass of K2P = no of moles * gram molar mass
= 0.0307*242.3 = 7.44g
dipotassium phthalate = 7.44g
How many grams of dipotassium phthalate (242.3 g/mol) must be added to 100 mL of 0.125...
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