Question

Physiological pH inside of living cells is about 7.4. The total free concentration of phosphate inside of cells is about 1 mM. Phosphate exists in four different protonation states.

Suppose that 100 ml of cellular extract is treated with 50 ml of 0.5 mM HCl. What is the final pH? [Assume that the HCl reacts only with the phosphate buffer in the cell extract.] Which form of phosphate is most abundant now?

Answer 1

According to the diagram, HPO₄^-2 (let x) and H₂PO₄^- (let y) are more abundant than the others at physiological pH. They together form a buffer.

given , x+y = 1 mM.....(1)

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

7.4 = 7.21 + log(x/y)

or, x/y = 1.55.....(2)

solving 1, and 2

x= 0.61 mM

y=0.39 mM

in 100 mL of extract,

no of moles of x = 0.061 mili moles

no of moles of y = 0.039 mili moles

50 mL of 0.5 mM HCl = 0.5*50/1000 = 0.025 mili moles of HCl

Hcl will react with x to form y.

new no of moles of x = x1 = 0.061 - 0.025 = 0.036 mili moles in 150 mL total solution

new no of moles of y = y1 = 0.039 + 0.025 = 0.064 mili moles in 150 mL total solution

So, [x1] = 0.036*1000/150 = 0.25 mM

[y1] = 0.064*1000/150 = 0.43 mM

pH = 7.21 + log 0.25/0.43 = 6.97

y1, i.e., H₂PO₄^- is most abundant now.