Determine the pH of each two-component solution. a. 0.0550 M in HI and 0.00850 M in HF b. 0.112 M in NaCl and 0.0953 M in KF c. 0.132 M in NH₄Cl and 0.150 M HNO₃ d. 0.0887 M in sodium benzoate and 0.225 M in potassium bromide e. 0.0450 M in HCl and 0.0225 M in HNO₃

Verified step by step guidance

1. Step 1: Identify the nature of each component in the solution. For example, HI is a strong acid, HF is a weak acid, NaCl is a neutral salt, KF is a salt that can affect pH due to the F- ion, NH4Cl is an acidic salt, HNO3 is a strong acid, sodium benzoate is a salt of a weak acid, and potassium bromide is a neutral salt.
2. Step 2: For solutions containing strong acids (like HI and HCl), calculate the pH directly from the concentration of the strong acid, as they dissociate completely in water. Use the formula: \( \text{pH} = -\log[\text{H}^+] \).
3. Step 3: For solutions containing weak acids (like HF), use the acid dissociation constant (Ka) to find the concentration of \( \text{H}^+ \) ions. Set up an equilibrium expression: \( \text{HF} \rightleftharpoons \text{H}^+ + \text{F}^- \) and solve for \( \text{H}^+ \) using the equation \( K_a = \frac{[\text{H}^+][\text{F}^-]}{[\text{HF}]} \).
4. Step 4: For solutions containing salts that can affect pH (like KF or NH4Cl), determine if the salt will hydrolyze in water. For KF, the F- ion can react with water to form HF and OH-, making the solution basic. For NH4Cl, the NH4+ ion can donate a proton to water, making the solution acidic.
5. Step 5: For mixtures of strong acids (like HCl and HNO3), add the concentrations of \( \text{H}^+ \) ions from each acid to find the total \( \text{H}^+ \) concentration, then calculate the pH using \( \text{pH} = -\log[\text{H}^+] \).