Question

5 (1) Calculate, stepwise, the weights of KH2PO4 and K2HPO4 . You need to prepare 1000 mL of 0.8 M phopsphate buffer, ph 7.5 (Two decimal place in all the calculation steps please).

(2) Calculate the K+ concentration of the phosphate buffer. And convert the final unit to mM

(3) if the pH of th phosphate buffer you made did not equal 7.5, can you explain why?

Answer 1

Q1

total V = 1000 mL = 1 L

total M = 0.8 M

total moles = MV = 1*0.8 = 0.8 moles of HPO4-2 and H2PO4-

HPO4-2 + H2PO4- =  0.8

if pH = 7.5

then, find ratio

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

7.50 = 7.21 + log((HPO4-2 / H2PO4-))

(HPO4-2 / H2PO4-) = 10^(7.50-7.21)= 1.9498

HPO4-2 = 1.9498* H2PO4-

from

HPO4-2 + H2PO4- =  0.8

1.9498* H2PO4-+ H2PO4- =  0.8

H2PO4- = 0.8/(2.9498) = 0.27120

HPO4-2 = 1.9498* 0.27120 = 0.5287

these are moles, we need mass

mol of KH2PO4 = 0.27120

mol of K2HPO4 =0.5287

mass of KH2PO4 = 0.27120* 136.0855= 36.906 g

mass of K2HPO4 =0.5287*174.2 = 92.099 g

then...

weigh = 92.099 g of K2HPO4 + 36.906 g of KH2PO4

add in a 1L beaker, all mass

then, add water up to V = 1000 mL mark

2)

mol of K+ = KH2PO4 +2*K2HPO4 = 0.27120+2*0.5287 = 1.3286 mol of K+

mmol = 1.3286*10^3 = 1328.6 mmol / L = 1328.6 mM

3)

this is most likely because of activity between salt and ions, the K+ presence will increase real interactions, i.e. this model is ideal and therefore, pH is not exactly 7.5