**Question:**What is the pH of a 0.555M solution of NaF?**Explanation:**This question requires determining the pH of a sodium fluoride (NaF) solution with a given molarity. NaF is a salt that completely dissociates in water to form sodium ions (Na⁺) and fluoride ions (F⁻).**Steps to Solution:**1. **Understand the Base Reaction:** NaF dissociates in water, allowing the fluoride ion (F⁻) to act as a weak base. The fluoride ion reacts with water to form hydrogen fluoride (HF) and hydroxide ions (OH⁻): \[ \text{F}^- + \text{H}_2\text{O} \rightarrow \text{HF} + \text{OH}^- \]2. **Determine Kb:** We need the base dissociation constant (Kb) for F⁻, which is related to the acid dissociation constant (Ka) of HF: \[ K_{\text{a}}(\text{HF}) \times K_{\text{b}}(\text{F}^-) = K_{\text{w}} \] \( K_{\text{w}} \) is the ion-product constant of water (\(1.0 \times 10^{-14} \, \text{at 25°C}\)). 3. **Calculate Hydroxide Concentration:** Use the Kb value and the initial concentration of F⁻ to find OH⁻ concentration. Use the formula: \[ [OH^-] = \sqrt{K_{\text{b}} \times [F^-]} \]4. **Convert to pH:** Calculate pOH from OH⁻ concentration: \[ \text{pOH} = -\log[OH^-] \] Then, convert to pH using: \[ \text{pH} = 14 - \text{pOH} \]These calculations will provide the pH of the 0.555M NaF solution.

In the given question we have to calculate the pH of salt solution, First of all, we have to identify the salt formation and then find out the ka of acid from the standard value and then using the equation of pH calculation of salt solution.here, we get the pH of the salt solution, pH = 8.46