At 27^ C .100 ~mL of 0.4 M HCl is mixed with 100 mL of 0.5 M NaOH solution. To the resultant solution. 800 mL of distilled water is added. What is the pH of final solution?

Step-by-step solution

Moles of HCl =0.4 ~m 0.1 ~L =0.04 moles For NaOH :- Moles of NaOH =0.5 M 0.1 ~L =0.05 moles HCl and NaOH react in 1: 1 Ratio HCl + NaOH NaCl + H ₂ O NaOH is in excess since 0.05 moles of NaOH react with 0.04 moles of HCl (0.05 moles -0.04 moles )=0.01 moles of NaOH remaining The total volume after mixing -100 ml of HCl +100 ml of NaOH +800 ml of distilled water 1000 ml or 1 L so the concentration of NaOH = 0.01 moles 1 I =0.01 ~m pOH =- [ OH ⁻ ] pOH =- [0.01]=2 As we know pH + pOH=14 pH =14- pOH =14-2=12