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 determine the concentration of Na ions in the buffer solution, we need to consider the dissociation of the salts Na2HPO4 and NaH2PO4 separately.
The dissociation reactions for Na2HPO4 and NaH2PO4 are as follows:
Na2HPO4 (s) ↔ 2Na+ (aq) + HPO4^2- (aq)
NaH2PO4 (s) ↔ Na+ (aq) + H2PO4^- (aq)
In the buffer solution, both salts are mixed in equal amounts, so we have an equal number of moles for Na2HPO4 and NaH2PO4.
Assume we have 1 mole of each salt originally. After dissociation, we would have:
For Na2HPO4:
1 mole of Na2HPO4 yields 2 moles of Na+ ions. Therefore, the concentration of Na+ ions from Na2HPO4 is 2 M.
For NaH2PO4:
1 mole of NaH2PO4 yields 1 mole of Na+ ion. Therefore, the concentration of Na+ ions from NaH2PO4 is 1 M.
Since these salts are mixed in equal amounts, the total concentration of Na+ ions in the buffer solution will be the sum of the Na+ concentrations from Na2HPO4 and NaH2PO4:
Total Na+ concentration = (concentration from Na2HPO4) + (concentration from NaH2PO4)
= 2 M + 1 M
= 3 M
Therefore, the concentration of Na+ ions in the buffer solution is 3 M, not 0.3 M as you initially mentioned.
To determine the concentration of Na ions in the buffer prepared from Na2HPo4 and NaH2PO4, you need to consider the dissociation of these salts in water.
First, let's write out the dissociation equations for Na2HPo4 and NaH2PO4:
Na2HPo4:
Na2HPo4 ⇌ 2Na+ + HPO4^2-
NaH2PO4:
NaH2PO4 ⇌ Na+ + H2PO4-
Now, let's analyze how these ions contribute to the concentration of Na ions in the buffer.
In the case of Na2HPo4, each formula unit dissociates to release 2 Na+ ions. Since the concentration of Na2HPo4 is the same as NaH2PO4 in the buffer mixture, the concentration of Na ions from Na2HPo4 is 2 * (0.1M) = 0.2M.
In the case of NaH2PO4, each formula unit dissociates to release 1 Na+ ion. Again, since the concentration of NaH2PO4 is the same as Na2HPo4 in the buffer mixture, the concentration of Na ions from NaH2PO4 is 1 * (0.1M) = 0.1M.
To find the total concentration of Na ions in the buffer, you simply add the contributions from Na2HPo4 and NaH2PO4:
Total concentration of Na ions = (Concentration from Na2HPo4) + (Concentration from NaH2PO4)
Total concentration of Na ions = 0.2M + 0.1M
Total concentration of Na ions = 0.3M
Therefore, the concentration of Na ions in the buffer is indeed 0.3M.