Question

A buffer is prepared by dissolving some NaH2PO4 in 50 ml of a 0.2513 M NaOH and diluting to a total volume of 100 ml. How many grams of NaH2PO4 are needed to give a buffer with a pH of 7.40? (MW NaH2PO4= 119.9772 g/mol)

Answer 1

NaH₂PO₄ will react with NaOH to form Na₂HPO₄

NaH₂PO₄ + NaOH Na₂HPO₄ + H₂O

Moles of NaOH = Moles of Na₂HPO₄ formed = 0.2513 x 50 = 12.565 mmol

Therefore the moles of NaH₂PO₄ consumed initially = 12.565 mmol

As the solution is diluted to 100 mL,

therefore, the molarity of Na₂HPO₄= 12.565/100 = 0.12565 M

pK_a of NaH₂PO₄ = 7.2

From Henderson-Hasselbach equation,

pH = pK_a + log [base]/[acid]

pH = pK_a + log [Na₂HPO₄]/[NaH₂PO₄]

or, 7.4 = 7.2 + log 0.12565/[NaH₂PO₄]

or, log 0.12565/[NaH₂PO₄] = 0.2

or, 0.12565/[NaH₂PO₄] = 1.58489

or, [NaH₂PO₄] = 0.12565/1.58489

or, [NaH₂PO₄] = 0.0793 M

or, moles of NaH₂PO₄ = 0.0793 x 100 = 7.93 mmoles

Therefore the total amount of NaH₂PO₄ added initially = (12.565 + 7.93) = 20.495 mmoles

                                                                                                                     = 0.020495 moles

MW of NaH₂PO₄ = 119.9772 g/mol

Therefore the quantity of NaH₂PO₄ added initially = 0.020495 x 119.9772 = 2.459 g