Calculate the pH of an aqueous solution containing 3.8 10-2 M HCl, 1.0 10-2 M H2SO4, and 3.3 10-2 M HCN.
HCl is 100% ionized; therefore, (H^+) = (HCl)
HCN is a weak acid
..........HCN ==> H^+ + CN^-
initial.0.033......0......0
change.....-x.......x.....x
equil.....0.33-x...x......x
Ka = (H^+)(CN^-)/(HCN)
Look up Ka, substitute it and the ICE chart values and solve for x = H^+, then convert to pH.
For H2SO4, see your other post just before this one.
To calculate the pH of the solution, we need to determine the concentration of the hydronium ion (H3O+). We can do this by considering the dissociation of each acid in the solution.
1. HCl dissociates as follows:
HCl → H+ + Cl-
2. H2SO4 dissociates as follows:
H2SO4 → 2H+ + SO4^2-
3. HCN dissociates as follows:
HCN → H+ + CN-
Given the concentrations of each acid, we can calculate the concentration of H+ for each acid using their dissociation equations.
1. For HCl, the concentration of H+ is 3.8 × 10^-2 M.
2. For H2SO4, the concentration of H+ is 2 × (1.0 × 10^-2 M) = 2 × 10^-2 M. (Since H2SO4 dissociates into 2H+ ions.)
3. For HCN, the concentration of H+ is 3.3 × 10^-2 M.
Now we need to calculate the total concentration of H+ ions in the solution by summing the individual concentrations:
Total [H+] concentration = [H+] from HCl + [H+] from H2SO4 + [H+] from HCN
= (3.8 × 10^-2) + (2 × 10^-2) + (3.3 × 10^-2)
= 9.1 × 10^-2 M
Lastly, to calculate the pH, we can use the relationship:
pH = -log[H+]
pH = -log(9.1 × 10^-2)
= -log(9.1) - log(10^-2)
= -log(9.1) + 2
= 1.04 + 2
= 3.04
Therefore, the pH of the solution is approximately 3.04.
To calculate the pH of the aqueous solution, you need to consider the dissociation of each acid present and determine the concentration of H+ ions.
1. HCl dissociates fully in water, so all of it will contribute to H+ concentration:
HCl → H+ + Cl-
The concentration of H+ ions is 3.8 × 10^-2 M.
2. H2SO4 is a strong acid and also fully dissociates in water:
H2SO4 → 2 H+ + SO4^2-
The concentration of H+ ions is 2 × (1.0 × 10^-2 M) = 2 × 10^-2 M.
3. HCN is a weak acid. To determine the concentration of H+ ions from this acid, you need to consider its acid dissociation constant, Ka. The balanced chemical equation for the dissociation of HCN is:
HCN ⇌ H+ + CN-
The Ka expression for this reaction is:
Ka = [H+][CN-] / [HCN]
Given that Ka = 6.2 × 10^-10 and the initial concentration [HCN] = 3.3 × 10^-2 M, you can assume that the initial concentration of H+ ions is negligible compared to the final concentration. This allows you to simplify the expression to:
Ka = [H+][CN-] / [HCN] ≈ [H+][CN-] / (initial concentration of HCN)
Rearranging the equation, you can solve for [H+]:
[H+] ≈ (Ka × [HCN]) / [CN-]
Since the acid dissociates in a 1:1 ratio, the concentration of CN- can be assumed to be equal to the concentration of HCN.
Therefore, [H+] ≈ Ka × [HCN]
[H+] ≈ 6.2 × 10^-10 × (3.3 × 10^-2 M) = 2.046 × 10^-11 M
4. Now, you have the concentration of H+ ions from all three acids:
- H+ ions from HCl: 3.8 × 10^-2 M
- H+ ions from H2SO4: 2 × 10^-2 M
- H+ ions from HCN: 2.046 × 10^-11 M
To calculate the total concentration of H+ ions, you need to sum up the contributions from each acid:
Total [H+] = (concentration of H+ from HCl) + (concentration of H+ from H2SO4) + (concentration of H+ from HCN)
Total [H+] = 3.8 × 10^-2 M + 2 × 10^-2 M + 2.046 × 10^-11 M
Finally, use the concentration of H+ ions to calculate the pH of the solution:
pH = -log10([H+])
Plug in the value for [H+] in the above equation to get the pH of the solution.