Question

prepare 250 ml of 0.5 m phosphate buffer of ph 7.6 give that pka = 7.21 using the equimolar method. to prepare the same, first prepare stock a 250 ml of 0.5 m k2hpo4 and stock b 250 ml of 0.5 m kh2po4. then calculate how much of these you would mix to prepare the final buffer. given: mw of k2hpo4 = 174.18; mw = of kh2po4 = 136.08

Answer 1

Solution.

$pH = pKa - lg(\frac{[KH2PO4]}{[K2HPO4]})$

$lg(\frac{[KH2PO4]}{[K2HPO4]}) = pKa - pH$

$\frac{[KH2PO4]}{[K2HPO4]} = 10^{pKa -pH}$

$\frac{[KH2PO4]}{[K2HPO4]} = 0.41$

V(KH2PO4, 0.5 М) = 72 mL

V(K2HPO4, 0.5 М) = 178 mL

Answer:

V(KH2PO4, 0.5 М) = 72 mL

V(K2HPO4, 0.5 М) = 178 mL