To calculate the pH of the solution obtained by mixing HCl and NaOH, we need to consider the neutralization reaction between the two compounds. The reaction between HCl (hydrochloric acid) and NaOH (sodium hydroxide) produces water (H₂O) and forms a salt (NaCl).
Given:
Volume of HCl solution (V₁) = 40 cm³
Concentration of HCl solution (C₁) = 0.2 M
Volume of NaOH solution (V₂) = 30 cm³
Concentration of NaOH solution (C₂) = 0.1 M
1. Determine the moles of HCl and NaOH used:
Moles of HCl = Concentration (C₁) × Volume (V₁)
Moles of HCl = 0.2 M × 0.04 L (converting cm³ to L)
Moles of HCl = 0.008 mol
Moles of NaOH = Concentration (C₂) × Volume (V₂)
Moles of NaOH = 0.1 M × 0.03 L (converting cm³ to L)
Moles of NaOH = 0.003 mol
2. Determine the limiting reagent:
The stoichiometry of the reaction between HCl and NaOH is 1:1, meaning that they react in a 1:1 ratio. Whichever reactant is present in a smaller amount will be the limiting reagent.
In this case, NaOH is present in a smaller amount (0.003 mol), which means it will be fully consumed during the reaction.
3. Determine the excess reagent and its remaining moles:
Since NaOH is the limiting reagent, we need to find the remaining moles of HCl.
Moles of HCl remaining = Moles of HCl initially - Moles of NaOH reacted
Moles of HCl remaining = 0.008 mol - 0.003 mol
Moles of HCl remaining = 0.005 mol
4. Calculate the concentration of HCl in the resulting solution:
Volume of resulting solution = Volume of HCl solution + Volume of NaOH solution
Volume of resulting solution = 0.04 L + 0.03 L
Volume of resulting solution = 0.07 L
Concentration of HCl in the resulting solution = Moles of HCl remaining / Volume of resulting solution
Concentration of HCl in the resulting solution = 0.005 mol / 0.07 L
Concentration of HCl in the resulting solution ≈ 0.071 M
5. Calculate the pH of the resulting solution:
pH = -log[H⁺]
pH = -log(0.071)
Using logarithm properties, we can determine the pH value:
pH ≈ -log(0.071)
pH ≈ -(-1.147)
pH ≈ 1.147
Therefore, the pH of the solution obtained by mixing 40 cm³ of 0.2 M HCl and 30 cm³ of 0.1 M NaOH is approximately 1.147.