Question

A buffer solution is composed of 1.126 g of kh2po4 and 5.672 g of Na2HPO4 . ( Ka for dihydrogen phosphate ion is 6.2x 10^-8.)

a. what is the PH of the buffer solution?

pH =

What mass of must be added to decrease the buffer solution pH by 0.10 unit from the value calculated in part a?

Mass = ____ g

Answer 1

kh2po4 --> K+ + H2PO4-

Na2HPO4 --> 2Na+ + HPO4-2

pKa = -log(6.2*10^-8) = 7.207

the buffer equation

pH = pKA2 + log(HPO4-2 / H2PO4-)

mol of kh2po4 = 136.086 g/mol

mol of KH2PO4 = 1.126/136.086 = 0.008274

mol of Na2HPO4 = 5.672/141.96 = 0.0399

the buffer:

pH = pKA2 + log(HPO4-2 / H2PO4-)

pH = 7.21 + log(H0.0399 / 0.008274) = 7.89325

find mass for:

pH drop = 0.1

i.e

pH = 7.89325 -0.1 = 7.79325

so

7.79325 = pKA2 + log(HPO4-2 / H2PO4-)

7.79325 = 7.21 + log(Na2HPO4 / KH2PO4 )

7.79325 = 7.21 + log(0.0399/ (0.008274+x))

solve for x

0.0399/ (0.008274+x)) = 10^(7.79325 -7.21)

0.0399/ 3.83 = 0.008274+x

x = 0.0399/ 3.83 -0.008274

x= 0.00214 mol of KH2PO4 must be added

mass = mol*MW = 0.00214 *136.086 = 0.291 g of KH2PO4