200ml of 0.15M H3AsO4 given Ka1=10^-3 Ka2=10^-7 Ka3=10^-12 50ml of 0.6M Na2HAsO4 and 100ml of 0.1M Naish are mixed. pH of solution is
What is Naish
Solution
To find the pH of the solution, we need to determine the concentration of H+ ions in the solution. We can do this by considering the dissociation of the weak acid H3AsO4 and the salt Na2HAsO4.
First, let's start by writing the dissociation equations for H3AsO4:
H3AsO4 ⇌ H+ + H2AsO4- (Ka1)
H2AsO4- ⇌ H+ + HAsO42- (Ka2)
HAsO42- ⇌ H+ + AsO43- (Ka3)
From the given Ka values, we can see that Ka1 > Ka2 > Ka3. This means that the first dissociation step (Ka1) will dominate, and the second and third steps can be neglected.
Next, let's calculate the number of moles of H3AsO4 originally present in the solution:
Number of moles of H3AsO4 = volume (in L) × molarity
= 0.200 L × 0.15 mol/L
= 0.030 mol
Since H3AsO4 is a weak acid, it only partially dissociates. Let x be the concentration of H+ ions (in mol/L) produced from the dissociation of H3AsO4. We can make the following assumptions:
[H+] = [H2AsO4-] ≈ x (from Ka1 expression)
[HAsO42-] ≈ 0 (since Ka2 and Ka3 can be neglected)
The initial concentration of H3AsO4 = [H3AsO4] = 0.15 M (given)
Using the equilibrium expression for Ka1: Ka1 = [H+][H2AsO4-] / [H3AsO4], we can substitute the approximate values:
10^-3 = x × x / 0.15
Simplifying, we find x^2 = 10^-3 × 0.15
x^2 = 1.5 × 10^-4
x ≈ 0.0122 mol/L
Since we mixed 50 mL of 0.6 M Na2HAsO4 and 100 mL of 0.1 M NaOH, we can calculate the total volume of the resulting solution:
Total volume = 50 mL + 100 mL = 150 mL = 0.150 L
Now, let's consider the concentration of H+ ions resulting from the dissociation of Na2HAsO4. Since Na2HAsO4 is a salt, it completely dissociates in water to produce Na+ and HAsO42- ions.
[HAsO42-] = [Na2HAsO4] = 0.6 M (given)
[H+] = [HAsO42-] ≈ 0.6 M
To find the total concentration of H+ ions in the solution, we add the contributions from H3AsO4 and Na2HAsO4:
Total [H+] = [H+] (from H3AsO4) + [H+] (from Na2HAsO4)
= 0.0122 M + 0.6 M
≈ 0.6122 M
Finally, we can calculate the pH of the solution using the formula:
pH = -log[H+]
pH = -log(0.6122)
pH ≈ 0.21
Therefore, the pH of the solution is approximately 0.21.