The Ka values of H3PO4 are shown below.
Ka1 7.5 10-3
Ka2 6.2 10-8
Ka3 4.8 10-13
What is the pH of a 0.19 M solution of NaH2PO4?
The pH at the first equivalence point of in the titration of H3PO4 is determined by the hydrolysis of the salt, NaH2PO4.
(H^+) = sqrt(k1k2).
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To find the pH of a 0.19 M solution of NaH2PO4, we need to consider the dissociation of NaH2PO4 in water and the subsequent dissociation of H2PO4-.
First, let's write the dissociation equation of NaH2PO4 in water:
NaH2PO4 ↔ Na+ + H2PO4-
Since NaH2PO4 is a salt of a weak acid (H3PO4), the H2PO4- ion will undergo further dissociation in water.
The equilibrium equation for the ionization of H2PO4- is as follows:
H2PO4- ↔ H+ + HPO4^2-
We need to determine the concentration of H+ ions in the solution to calculate the pH.
Let's define the following variables:
[H2PO4-] = Concentration of H2PO4-
[H+] = Concentration of H+ ions
At equilibrium, the equation for Ka2 can be written as:
Ka2 = ([H+][HPO4^2-])/[H2PO4-]
Given that the initial concentration of NaH2PO4 is 0.19 M, and assuming complete dissociation, the concentration of H2PO4- is also 0.19 M.
Thus, we have:
[H2PO4-] = 0.19 M
Substituting the given Ka2 value and [H2PO4-] into the equation, we can solve for [H+] concentration.
6.2 × 10^-8 = ([H+][HPO4^2-])/0.19
Simplifying the equation, we get:
[H+] = (6.2 × 10^-8 × 0.19)/1
[H+] = 1.178 × 10^-8 M
Since [H+] represents the concentration of H+ ions in the solution, we can use this value to calculate the pH.
The pH is calculated using the formula:
pH = -log[H+]
pH = -log(1.178 × 10^-8)
pH ≈ 7.93
Therefore, the pH of a 0.19 M solution of NaH2PO4 is approximately 7.93.
To find the pH of a 0.19 M solution of NaH2PO4, we need to consider the dissociation of the H2PO4- ion and the subsequent formation of H3O+ ions.
NaH2PO4 is a salt that dissociates in water, producing H2PO4- ions and Na+ ions. The H2PO4- ions can then react with water to form H3O+ ions:
H2PO4- + H2O ⇌ H3O+ + HPO42-
Since we are given the Ka values for H3PO4, we can assume that the first dissociation is the main one. Therefore, we can consider the equilibrium between H2PO4- and H3O+ to calculate the pH.
The Ka value for the first dissociation (Ka1) is given as 7.5x10^-3. This means that:
Ka1 = [H3O+][HPO42-] / [H2PO4-]
Assuming that x is the concentration of H3O+ and HPO42-, and since the initial concentration of H2PO4- is 0.19 M, we can write:
x^2 / (0.19 -x) = 7.5x10^-3
Simplifying this equation yields a quadratic equation, which can be solved to find the concentration of H3O+.
Once the concentration of H3O+ is known, we can calculate the pH using the equation:
pH = -log[H3O+]
Calculating this value will give us the pH of the 0.19 M solution of NaH2PO4.