Question

# How many grams of dipotassium phthalate (242.3 g/mol) must be added to 100 mL of 0.125...

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

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