A buffer prepared by mixing equal amounts of the salts Na2HPo4 and NaH2PO4 to give a total phosphate concentration of 0.1M. What is the concentration of Na ions?
I believe it is 0.3M but am having trouble understanding why when I write out the dissociation of the Na2HPo4 and NaH2PO4. Thanks
To understand the concentration of Na ions in the buffer solution, we need to consider the dissociation of the salts Na2HPO4 and NaH2PO4 in water.
Na2HPO4 dissociates into 2 Na+ ions and 1 HPO4^2- ion:
Na2HPO4 (s) ↔ 2 Na+ (aq) + HPO4^2- (aq)
NaH2PO4 dissociates into 1 Na+ ion, 1 H+ ion, and 1 H2PO4^- ion:
NaH2PO4 (s) ↔ Na+ (aq) + H+ (aq) + H2PO4^- (aq)
Since the buffer solution is prepared by mixing equal amounts of these two salts, the concentration of each salt will be 0.05 M (half of the total phosphate concentration of 0.1 M). Therefore, the concentration of Na+ ions in the buffer solution is the sum of the concentrations of Na+ ions from both salts.
For Na2HPO4, the concentration of Na+ ions is 2 times the concentration of Na2HPO4:
Concentration of Na+ ions from Na2HPO4 = 2 * 0.05 M = 0.1 M
For NaH2PO4, the concentration of Na+ ions is 1 times the concentration of NaH2PO4:
Concentration of Na+ ions from NaH2PO4 = 1 * 0.05 M = 0.05 M
Adding these two concentrations gives us the total concentration of Na+ ions in the buffer solution:
Concentration of Na+ ions = 0.1 M + 0.05 M = 0.15 M
Therefore, the concentration of Na+ ions in the buffer solution is 0.15 M, not 0.3 M.