Answer:
To solve this problem, we'll use the concept of stoichiometry and the volume of solutions involved in the titration.
Given:
1. Volume of acid (HCl) = 25 cm³
2. Molar concentration of base (NaOH) = 0.15 mol/dm³
3. Average volume of base used for total acid = 29.95 cm³
We'll start by calculating the moles of acid (HCl) involved in the mixture.
i. The volume of NaOH required for complete neutralization of HCl:
Since the reaction between NaOH and HCl is 1:1 according to the equation:
NaOH + HCl → NaCl + H2O
We'll use the equation:
\[ \text{moles} = \text{concentration} \times \text{volume} \]
\[ \text{moles of HCl} = \text{concentration of HCl} \times \text{volume of HCl} \]
\[ \text{moles of HCl} = (0.15 \, \text{mol/dm³}) \times (25/1000 \, \text{dm³}) \]
\[ \text{moles of HCl} = 0.15 \times 0.025 \]
\[ \text{moles of HCl} = 0.00375 \, \text{mol} \]
Since the reaction is 1:1, the moles of NaOH required for complete neutralization of HCl will also be 0.00375 mol.
ii. The volume of NaOH required for complete neutralization of the total acid:
We already know that the average volume of NaOH used for total acid is 29.95 cm³. This volume is slightly larger than the initial 25 cm³ of HCl due to the additional volume needed to neutralize the second acid (CH3COOH). Therefore, the volume of NaOH needed for total acid is 29.95 cm³.
iii. The volume of NaOH required for complete neutralization of CH3COOH:
Since the reaction between NaOH and CH3COOH is also 1:1 according to the equation:
NaOH + CH3COOH → CH3COONa + H2O
We'll use the same concept as above. However, we'll need additional information about the concentration and volume of CH3COOH to calculate this. If you provide those details, I can help you further with part iii.
