what is the pH of a solution created by adding 15.5 g of KClO to 228 mL of 1.08 M HClO(aq).?
To determine the pH of a solution, we need to consider the reaction between the substances involved. In this case, potassium chlorate (KClO) and hydrochloric acid (HClO) will react to form a solution.
To calculate the pH, we can follow these steps:
1. Determine the moles of KClO:
- First, find the molar mass of KClO, which is 74.55 g/mol (potassium: 39.10 g/mol, chlorine: 35.45 g/mol, oxygen: 16.00 g/mol).
- Divide the mass of KClO (15.5 g) by the molar mass to get the number of moles: 15.5 g / 74.55 g/mol = 0.2081 mol.
2. Determine the moles of HClO:
- Volume (V) of HClO solution = 228 mL = 0.228 L.
- Multiply the volume by the molarity (1.08 M) to get the number of moles: 1.08 mol/L * 0.228 L = 0.24624 mol.
3. Determine the limiting reactant:
- Compare the moles of KClO and HClO to find the limiting reactant (the one that is completely consumed during the reaction).
- In this case, it is clear that HClO is the limiting reactant because there is an excess of KClO.
4. Calculate the moles of HCl produced:
- Since the reaction is 1:1, the moles of HClO that reacted are equal to the moles of HCl produced. In this case, it is 0.24624 mol.
5. Calculate the concentration of HCl:
- Divide the moles of HCl by the new total volume of the solution.
- The new total volume is the sum of the initial volume (0.228 L) and the final volume of HCl produced by the reaction (it does not contribute to the volume, so let's assume it is negligible).
- The final volume can be approximated as 0.228 L (no significant volume change).
- Concentration (C) of HCl = moles of HCl (0.24624 mol) / total volume (0.228 L) = 1.079 M.
6. Calculate the pH:
- The pH is determined using the equation pH = -log10[H+].
- Since HCl is a strong acid, it dissociates completely, and the concentration of H+ ions is equal to the concentration of HCl.
- Plug in the concentration of HCl (1.079 M) into the equation to find the pH: pH = -log10(1.079) ≈ 0.97.
Therefore, the pH of the solution created by adding 15.5 g of KClO to 228 mL of 1.08 M HClO(aq) is approximately 0.97.