Question

In: Chemistry

A buffer is prepared by dissolving some NaH2PO4 in 50 ml of a 0.2513 M NaOH...

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)

Solutions

Expert Solution

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

pKa of NaH2PO4 = 7.2

From Henderson-Hasselbach equation,

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

pH = pKa + 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


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