Calculate the pH of a 0.10 M NaHX salt solution. Ka (H2X) = 2.1 x 10-2 Ka (HX-) = 4.5 x 10-8
Can someone explain the steps to do this please?
The pH will be determined by the hydrolysis of the salt, NaX.
...........X^- + HOH ==> HX + OH^-
initial..0.10.............0....0
change....-x..............x.....x
equil....0.10-x...........x.....x
Kb for the salt = (Kw/Ka) = (x)(x)/(0.10-x)
Kw = 1E-14
Ka is Ka for HX, whatever that is. Solve for x = (OH^-), then convert to H^+ and to pH.
Sure! Here are the step-by-step instructions to calculate the pH of a 0.10 M NaHX salt solution:
Step 1: Write out the balanced chemical equations:
NaHX (aq) ⇄ Na+ (aq) + HX- (aq)
Step 2: Determine the initial concentration of the HX- ion. In this case, it is given as 0.10 M.
Step 3: Define the equilibrium constant expression for the dissociation of HX-. The equilibrium constant (Ka) for the dissociation reactions of H2X and HX- are given as 2.1 x 10^-2 and 4.5 x 10^-8, respectively.
For the dissociation of HX-:
Ka = [H+][X-] / [HX-]
Step 4: Since HX- is a monoprotic acid, the concentration of [H+] ions will be equal to the concentration of [HX-] ions at equilibrium.
Therefore, we can simplify the equation as:
[H+]^2 / [HX-] = Ka
Step 5: Substitute the given values into the equation:
[H+]^2 / 0.10 M = 4.5 x 10^-8
Step 6: Rearrange the equation to solve for [H+]:
[H+]^2 = (4.5 x 10^-8) * (0.10 M)
[H+]^2 = 4.5 x 10^-9
[H+] = √(4.5 x 10^-9)
[H+] ≈ 2.12 x 10^-5 M
Step 7: Calculate the pH using the formula:
pH = -log[H+]
pH = -log(2.12 x 10^-5)
pH ≈ 4.67
Therefore, the pH of the 0.10 M NaHX salt solution is approximately 4.67.
To calculate the pH of a solution containing a salt, you need to consider the dissociation of the salt in water and determine the concentration of the hydrogen ions (H+) in the solution.
Step 1: Write the dissociation reaction of the salt in water. In this case, NaHX will dissociate into Na+ and HX- ions.
NaHX ⇌ Na+ + HX-
Step 2: Identify the relevant acid dissociation constants (Ka) and write the dissociation reactions for both H2X and HX-.
H2X ⇌ H+ + X- (Ka = 2.1 x 10^-2)
HX- ⇌ H+ + X- (Ka = 4.5 x 10^-8)
Step 3: Determine the concentration of H+ ions in the solution. Since H2X and HX- will dissociate to form H+ ions, we need to consider both reactions.
Let's assume the concentration of HX- ions that dissociate is x M. Therefore, the concentration of H+ ions from HX- is also x M.
From the second dissociation reaction, the concentration of H+ ions is also x M.
Therefore, the total concentration of H+ ions in the solution is 2x M.
Step 4: Write the equation for the Ka expression for both dissociation reactions.
Ka = [H+][X-] / [HX-]
For the dissociation of H2X, we can assume that the reactant concentration is unchanged since H2X is a solid salt.
Ka (H2X) = [H+][X-] / [H2X] = [H+][X-] / (0.10 M)
Therefore, [H+][X-] = (Ka (H2X)) * (0.10 M)
Similarly, for the dissociation of HX-, we can assume that the reactant concentration is x M.
Ka (HX-) = [H+][X-] / [HX-] = (x)(x) / (0.10 - x) M
Step 5: Solve the quadratic equation to find the value of x.
Using the value of Ka (HX-) = 4.5 x 10^-8, we can substitute the values into the equation and solve for x.
4.5 x 10^-8 = (x)(x) / (0.10 - x) M
Step 6: Once you find the value of x, substitute it back into the equation for [H+] to find the concentration of hydrogen ions.
[H+] = 2x M
Step 7: Calculate the pH using the concentration of H+ ions.
pH = -log[H+]
You can now calculate the pH using the concentration of H+ ions determined in Step 6.