Calculate the pH of a solution that is made up to contain the following analytical concentrations : 0.05 M in H3PO4 and 0.02 M in NaH2PO4
To calculate the pH of this solution, you need to know the dissociation constants (Ka values) for the acid H3PO4. The dissociation of H3PO4 in water can be represented by the following equations:
H3PO4 ⇌ H+ + H2PO4- (1st dissociation)
H2PO4- ⇌ H+ + HPO4^2- (2nd dissociation)
HPO4^2- ⇌ H+ + PO4^3- (3rd dissociation)
For simplicity, we can assume that the second and third dissociation constants are so small that we can ignore them. This assumption is valid because phosphoric acid is a weak acid and does not dissociate fully in water.
The value of the first dissociation constant (Ka1) for H3PO4 is 7.5 x 10^-3.
Now, let's calculate the concentrations of H+ and the corresponding pH:
Step 1: Calculate the concentration of H+ ions from the dissociation of H3PO4.
Since H3PO4 dissociates in a 1:1 ratio, the concentration of H+ ions will be equal to the concentration of H3PO4.
[H+] = 0.05 M (concentration of H3PO4)
Step 2: Calculate the pH using the equation: pH = -log[H+]
pH = -log(0.05)
pH ≈ 1.30
Therefore, the pH of the solution containing 0.05 M H3PO4 and 0.02 M NaH2PO4 is approximately 1.30.