Question

How many grams of dipotassium succinate trihydrate (K₂C₄H₄O₄·3H₂O, MW = 248.32 g/mol) must be added to 650.0 mL of a 0.0420 M succinic acid solution to produce a pH of 5.867? Succinic acid has pK_a values of 4.207 (pK_a1) and 5.636 (pK_a2).

Answer 1

Desved pH of buffer 2 5.867 pkg of succinic acid = 5.636 Now, dipottasium succinate ic K₂ Cu Hu Oy and succinic acid (C ₂ H 6 Oy) are available pka, is equali dissociation constant for Cu HgOu Cu Hy Oy² +H+ So, festly mono succinate (Cy Hg Oy) must be available . Cu HG Ou + Cu Hu Oy - 24 H5 On (succinic acid) (a disuccinate) (monosuccinate) It indicates, equal moles of succinic acid and disuccinate form double moles of mono succinate Moles of succinic acid present a 0.0420 x 650 = 0.0273 mol.

moles of from mono succinate that can be founed available succinic acid a 2x0.0273 = 0.0546 mol For this polasium disuccinate required 2/0 0273 mol. Til Now, we have 0.0546 mol of mono succinate and no other species - Using Henderson - Hasselbalch equation! PH = pka, & log (moles of disuccinate) (moles of moniccinate) 5.867 = 5.636 + log (moles of disuccinate) (0.0546 mol) 5.867 - 5.636 = log (moles of disuccinate) 0.0546 mol 108.231 = moles of disuccinate 0.0546 mol Required moles of disuccinate a 1.702 XO 10546 20.0929 moll Total moles of disuccinate to be adeleda 0.0929+ 0.0273 20-1202 was mol

Molar mass of potassium disuccinate = 248.32 g/mol

Moles of potassium disuccinate required = 0.1202 mol

Mass of potassium disuccinate required

= 0.1202 mol*248.32g/mol

= 29.848 g