Question

How many grams of NaH2PO4 (119.96 g/mol) are needed to make 500 mL of a 0.2 M phosphate buffer at pH 7.5? The pKa of phosphate is 7.21

Answer 1

no of moles of Phosphate   = molarity * volume in L

                                             = 0.2*0.5   = 0.1 moles

PH   = Pka + log[Phosphate]/[NaH2PO4]

   7.5   = 7.21 + log0.1/[NaH2PO4]

log0.1/[NaH2PO4]   = 7.5-7.21

log0.1/[NaH2PO4]     = 0.29

0.1/[NaH2PO4]         = 10^0.29

0.1/[NaH2PO4]       = 1.9498

[NaH2PO4]            = 0.1/1.9498   = 0.0512moles

mass of NaH2PO4 = no of moles* gram molar mass

                                   = 0.0512*119.96   =6.142g >>> answer