The triprotic acid H3A has ionization constants of Ka1 = 9.9× 10–2, Ka2 = 2.7× 10–8, and Ka3 = 5.9× 10–11.
a). Calculate the following values for a 0.0570 M solution of NaH2A
H+ ??
[H2A^-/H3A]??
b). Calculate the following values for a 0.0570 M solution of Na2HA.
H+ ??
[HA^2-/H2A]??
they give you the formulas to solve for H+ but i've tried it so many times and it is still wrong..
Let's try using the formula and see what we get? Or better still post the formula and your work and I can find the error.
To solve this problem, we need to apply the principles of acid-base equilibria and use the given ionization constants (Ka values) of the triprotic acid H3A. Let's break down the steps to find the answers:
a) For a 0.0570 M solution of NaH2A:
Step 1: Write the balanced equation for the ionization of H3A into H+ ions and its conjugate base H2A-.
H3A ⇌ H+ + H2A-
Step 2: Set up an ICE table to represent the initial, change, and equilibrium concentrations of each species.
Initial concentration:
[H3A] = 0.0570 M
[H+] = 0 M
[H2A-] = 0 M
Change in concentration:
[H3A] decreases by x
[H+] increases by x
[H2A-] increases by x
Equilibrium concentration:
[H3A] = 0.0570 - x
[H+] = x
[H2A-] = x
Step 3: Write the expression for the equilibrium constant (Ka1) for the first ionization:
Ka1 = [H+][H2A-]/[H3A]
Substitute the equilibrium concentrations into the expression:
Ka1 = (x)(x)/(0.0570 - x)
Step 4: Apply the approximation: Since x is small compared to 0.0570, we can assume that (0.0570 - x) ≈ 0.0570 in the denominator.
Now the equation becomes:
Ka1 = x^2/0.0570
Step 5: Substitute the given value of Ka1 (9.9 × 10^–2) into the equation and solve for x:
9.9 × 10^–2 = x^2/0.0570
x^2 = 0.0570 × 9.9 × 10^–2
x = √(0.0570 × 9.9 × 10^–2)
x ≈ 0.0730
Step 6: Calculate the concentration of H+ and [H2A-]:
[H+] = x ≈ 0.0730 M
[H2A-] = x ≈ 0.0730 M
b) For a 0.0570 M solution of Na2HA:
Step 1: Write the balanced equation for the ionization of H2A- into H+ ions and its conjugate base HA2-.
H2A- ⇌ H+ + HA2-
Step 2: Set up an ICE table to represent the initial, change, and equilibrium concentrations of each species.
Initial concentration:
[HA2-] = 0.0570 M
[H+] = 0 M
[H2A-] = 0 M
Change in concentration:
[HA2-] decreases by x
[H+] increases by x
[H2A-] increases by x
Equilibrium concentration:
[HA2-] = 0.0570 - x
[H+] = x
[H2A-] = x
Step 3: Write the expression for the equilibrium constant (Ka2) for the second ionization:
Ka2 = [H+][HA2-]/[H2A-]
Substitute the equilibrium concentrations into the expression:
Ka2 = (x)(0.0570 - x)/x
Step 4: Apply the approximation: Since x is small compared to 0.0570, we can assume that (0.0570 - x) ≈ 0.0570 in the denominator.
Now the equation becomes:
Ka2 = (x)(0.0570 - x)/0.0570
Step 5: Substitute the given value of Ka2 (2.7 × 10^–8) into the equation and solve for x:
2.7 × 10^–8 = (x)(0.0570 - x)/0.0570
Step 6: This equation is quadratic, so it needs to be solved using the quadratic formula. The final result for x will be small, indicating that the approximation was valid:
Now, using the quadratic formula, we find:
x ≈ 2.3 × 10^–5
Step 7: Calculate the concentration of H+ and [HA2-]:
[H+] = x ≈ 2.3 × 10^–5 M
[HA2-] = 0.0570 - x ≈ 0.0570 M
By following these steps, you should be able to find the correct values for H+ and [H2A-]/[HA2-] in both cases. Make sure to double-check your calculations to avoid any errors.