Question

The pKa of phosphoric acid: 7.21. FW of Salt: 141.96. FW of acid 137.99. Make 0.5 L of a 0.03M sodium phosphate, ph 7.3 buffer.

Answer 1

total no of mol of buffer = M*V   =   0.5*0.03 = 0.015 mole

let us take , no of mole of salt = x mole

             no of mole of acid = 0.015-x mole

pH of acidicbuffer = pka + log(salt/acid)

7.3 = 7.21+log(x/(0.015-x))

x = 0.00827 mole

no of mole of salt = x = 0.00827 mole

             no of mole of acid = 0.015-x   = 0.015- 0.00827 = 0.00673mole

amount of salt must take = n*Mwt = 0.00827*141.96 = 1.174 g

amount of acid must take = n*Mwt = 0.00673*137.99 = 0.93 g

so that , take both salt,acid in a beaker and add water upto 0.5 L.so that buffer is prepared.