Physiological pH inside of living cells is about 7.4. The total free concentrati
ID: 979550 • Letter: P
Question
Physiological pH inside of living cells is about 7.4. The total free concentration of phosphate inside of cells is about 1mM. Phosphate exists in four different protonation states.A) Which form of phosphate is the most abundant at physiological pH?
B) Suppose that 100 ml of cellular extract is treated with 50 ml of 0.5 mM HCl. (Assume that the HCl reacts only with the phosphate buffer in the cell extract.) Which form of the phosphate is most abundant now? Physiological pH inside of living cells is about 7.4. The total free concentration of phosphate inside of cells is about 1mM. Phosphate exists in four different protonation states.
A) Which form of phosphate is the most abundant at physiological pH?
B) Suppose that 100 ml of cellular extract is treated with 50 ml of 0.5 mM HCl. (Assume that the HCl reacts only with the phosphate buffer in the cell extract.) Which form of the phosphate is most abundant now?
A) Which form of phosphate is the most abundant at physiological pH?
B) Suppose that 100 ml of cellular extract is treated with 50 ml of 0.5 mM HCl. (Assume that the HCl reacts only with the phosphate buffer in the cell extract.) Which form of the phosphate is most abundant now?
Explanation / Answer
Physiological pH inside of living cells is about 7.4. The total free concentration of phosphate inside of cells is about 1mM. Phosphate exists in four different protonation states.
A) Which form of phosphate is the most abundant at physiological pH?
Apply Buffer equation
pH = pKa2 + log(HPO4-2/H2PO4-)
7.40 = 7.21 + log(HPO4-2/H2PO4-)
(HPO4-2/H2PO4-) = 10^(7.40-7.21)
(HPO4-2/H2PO4-) = 1.54881
Clearly the LEAST acidic, i.e. the conjugate base is present in a larger extent, i.e. HPO4-2 > H2PO4-
B) Suppose that 100 ml of cellular extract is treated with 50 ml of 0.5 mM HCl. (Assume that the HCl reacts only with the phosphate buffer in the cell extract.) Which form of the phosphate is most abundant now?
First, calculate total H+ added from HCl --> mmol = MV = 50*0.5 = 25 mmol of H+
Now; this is plenty of H+
This will potonate:
HPO4-2 + H+(aq) <--> H2PO4-
Therefre, H2PO4- increases; HPO4- decreases
the acidic state of HPO4- will increase