find the pH of each of the following solutions of mixtures of acids: .125M in HBr and .130 M in HCHO2, .155M in HNO2 and 9.0*10^-2M in HNO3, 5.5×10−2 in acetic acid and 5.5×10−2 in hydrocyanic acid

For 1 and 2, calculate the H^+ from the strong acid first, use the H^+ from that calculation as a common ion to determine the H^+ from the weak acid, add them together (I suspect the weak acid will contribute so little it won't make much difference) for the total H^+. For the third one, follow the same procedure except now you are dealing with two weak acid BUT one is much stronger than the toher.

To find the pH of each of the given solutions, we need to use the concept of acid dissociation and understand the equilibrium between the acid and its conjugate base.

1. Solution of HBr and HCHO2:
First, we need to determine the Ka (acid dissociation constant) values for both acids:
- HBr does not dissociate significantly in water, so we can assume it is a strong acid and fully ionizes. Therefore, we can neglect its contribution to the overall pH of the solution.
- HCHO2 is a weak acid, so we need to consider its dissociation. The Ka value for HCHO2 is 1.8 × 10^-4.

Now, let's calculate the pH of the solution with HCHO2 alone:
- Use the Ka value to find the concentration of H+ ions in the solution.
- For a solution of 0.130 M in HCHO2, we can assume, initially, that all of it is undissociated. Therefore, the initial concentration of HCHO2 is 0.130 M.
- Since HCHO2 is a weak acid, we can assume that some of it dissociates into H+ and CHO2- ions. Let's say x is the concentration of H+ ions formed.
- At equilibrium, the concentration of HCHO2 will be (0.130 - x), and the concentration of H+ ions will be x.
- Writing the equilibrium expression for HCHO2 as HCHO2 ⇌ H+ + CHO2-, we have:
Ka = [H+][CHO2-] / [HCHO2]

- Substitute the values:
1.8 × 10^-4 = x * x / (0.130 - x)

- The above equation is quadratic, so we need to solve it to find the value of x (concentration of H+ ions).
- Once x is determined, we can calculate the pH using the formula pH = -log10([H+]).

2. Solution of HNO2 and HNO3:
Similar to the previous solution, let's consider each acid separately.
- HNO2 is a weak acid, with a Ka value of 4.5 × 10^-4.
- HNO3 is a strong acid, so we can neglect its contribution to the pH.

To find the pH of the solution with HNO2 alone:
- Assuming the 0.155 M HNO2 is undissociated initially, the concentration of HNO2 is 0.155 M.
- Let x be the concentration of H+ ions formed during the dissociation.
- At equilibrium, the concentration of HNO2 will be (0.155 - x), and the concentration of H+ ions will be x.
- The equilibrium expression for HNO2 is HNO2 ⇌ H+ + NO2-.
- The equilibrium equation becomes 4.5 × 10^-4 = x * x / (0.155 - x).
- Solve the equation to find the concentration of H+ ions (x) and calculate the pH using pH = -log10([H+]).

3. Solution of acetic acid and hydrocyanic acid:
Again, consider each acid individually.
- Acetic acid (CH3COOH) is a weak acid with a Ka value of 1.8 × 10^-5.
- Hydrocyanic acid (HCN) is also a weak acid with a Ka value of 6.2 × 10^-10.

To find the pH of the solution with acetic acid alone:
- Assuming the 5.5 × 10^-2 M acetic acid is initially undissociated, the concentration of acetic acid is 5.5 × 10^-2 M.
- Let x be the concentration of H+ ions formed during the dissociation.
- At equilibrium, the concentration of acetic acid will be (5.5 × 10^-2 - x), and the concentration of H+ ions will be x.
- The equilibrium expression for acetic acid is CH3COOH ⇌ H+ + CH3COO-.
- The equilibrium equation becomes 1.8 × 10^-5 = x * x / (5.5 × 10^-2 - x).
- Solve the equation to find the concentration of H+ ions (x) and calculate the pH using pH = -log10([H+]).

Repeat the same steps for the hydrocyanic acid to find its pH.

Note: It is important to use an appropriate calculator or software to solve the quadratic equations obtained from the equilibrium expressions and calculate the final pH values for each solution.