The pH of a solution made by adding 0.001 mole of NaOH to 100cm3of a s...
Calculation of pH of the solution:
- The reaction between CH3COOH and NaOH can be represented as follows:
CH3COOH + NaOH → CH3COONa + H2O
- The initial concentration of CH3COOH is 0.50 M, and the concentration of CH3COONa is also 0.50 M as it is a salt of CH3COOH.
- Since NaOH is a strong base, it will completely react with CH3COOH to form CH3COONa and water.
- 0.001 mole of NaOH is added to the solution, which means 0.001 mole of CH3COOH will be neutralized.
- The final concentration of CH3COOH after neutralization will be 0.50 M - 0.001 M = 0.499 M.
Calculating the pH:
- The equilibrium constant expression for the dissociation of CH3COOH is Ka = [CH3COO-][H3O+]/[CH3COOH].
- Using the given pKa value of CH3COOH (4.744), we can calculate the Ka value as 10^(-4.744).
- Now, we can calculate the concentration of H3O+ ions using the Ka expression and the concentration of CH3COOH.
- Finally, pH is calculated using the formula pH = -log[H3O+].
Result:
- After the above calculations, the pH of the solution is found to be 4.76.
Therefore, the pH of the solution made by adding 0.001 mole of NaOH to 100cm3 of a solution which is 0.50 M in CH3COOH and 0.50 M in CH3COONa is 4.76.