What is the hydronium ion concentration of a buffer solution that is 0.10M NaHPO4 and 0.50M NaH2PO4? For H3PO4,Ka1=6.9x10^-3, Ka2=6.2x10^-8, Ka3=4.8x10^-13
Use the Henderson-Hasselbalch equation.
Does it matter which Ka I use?
Nvm got it ..thanks
To find the hydronium ion concentration of the buffer solution, we need to consider the dissociation of NaH2PO4 and NaHPO4 in water.
First, let's write the dissociation reactions for NaH2PO4 and NaHPO4:
NaH2PO4 ↔ Na+ + H2PO4-
NaHPO4 ↔ Na+ + HPO42-
Since these are weak acid and its salt, we also need to consider the dissociation of H2PO4- and HPO42- in water:
H2PO4- ↔ H+ + HPO42-
HPO42- ↔ H+ + PO43-
Now let's consider the ionization constants (Ka) of H3PO4 to calculate the concentration of H+ in solution.
Ka1 = [H+][H2PO4-] / [H3PO4]
Ka2 = [H+][HPO42-] / [H2PO4-]
Ka3 = [H+][PO43-] / [HPO42-]
Given the values of Ka1, Ka2, and Ka3, we can assume that the dissociation of NaH2PO4 and NaHPO4 primarily occurs due to the common ion effect.
Since the concentration of H3PO4 isn't given, we can assume it's negligibly small compared to the concentration of NaH2PO4 and NaHPO4. Therefore, we can assume that the concentration of H2PO4- (from NaH2PO4) and HPO42- (from NaHPO4) are equal.
Let's denote the equilibrium concentrations of H2PO4- and HPO42- as [H2PO4-]eq and [HPO42-]eq, respectively.
From the dissociation reactions of NaH2PO4 and NaHPO4, we know that [H2PO4-]eq = [NaH2PO4] and [HPO42-]eq = [NaHPO4]. Given the concentrations of NaH2PO4 and NaHPO4 as 0.10 M and 0.50 M, respectively, we have [H2PO4-]eq = 0.10 M and [HPO42-]eq = 0.50 M.
Now, using the equilibrium expressions for Ka2 and Ka3, we can express [H+] in terms of [H2PO4-]eq and [HPO42-]eq:
For Ka2: 6.2x10^-8 = [H+][HPO42-] / [H2PO4-]
Simplifying, we get [H+] = 6.2x10^-8 * [H2PO4-] / [HPO42-]
For Ka3: 4.8x10^-13 = [H+][PO43-] / [HPO42-]
Simplifying, we get [H+] = 4.8x10^-13 * [HPO42-] / [PO43-]
Since we assumed that [H2PO4-]eq = [HPO42-]eq, we can set up an equation by equating the [H+] expressions:
6.2x10^-8 * 0.10 / [HPO42-]eq = 4.8x10^-13 * [HPO42-]eq / [PO43-]
Simplifying further, we find:
0.62 * 10^-8 = 4.8x10^-13 / [PO43-]
Rearranging the equation, we find that [PO43-] = 4.8x10^-13 / (0.62 * 10^-8)
Using this calculated value of [PO43-], we can substitute it back into the expression for [H+] in the dissociation of HPO42- to find the hydronium ion concentration.
[H+] = 4.8x10^-13 * [HPO42-]eq / [PO43-]
= 4.8x10^-13 * [HPO42-]eq / (4.8x10^-13 / (0.62 * 10^-8))
= 0.62 * 10^-8 * [HPO42-]eq
Since we know that [H2PO4-]eq = [HPO42-]eq = 0.10 M, we have:
[H+] = 0.62 * 10^-8 * 0.10
= 6.2 * 10^-9 M
Therefore, the hydronium ion concentration of the buffer solution is 6.2 * 10^-9 M.