On adding of naoh to ch3cooh solution ,60% of the acid is neutralize.i...
pH of the resulting solution after adding NaOH to CH3COOH solution
Introduction:
When NaOH is added to a solution of CH3COOH (acetic acid), a neutralization reaction takes place. The reaction can be represented as follows:
CH3COOH + NaOH → CH3COONa + H2O
The reaction shows that one molecule of CH3COOH reacts with one molecule of NaOH to form one molecule of CH3COONa (sodium acetate) and one molecule of water.
Given:
60% of the acetic acid is neutralized after adding NaOH.
pKa of CH3COOH = 4.7
Calculating the concentration of the remaining acetic acid:
Since 60% of the acetic acid is neutralized, it means that 40% of the acid remains unreacted.
Therefore, if the initial concentration of acetic acid is denoted as [CH3COOH]0, the concentration of the remaining acetic acid is:
[CH3COOH]remaining = 0.4 × [CH3COOH]0
Calculating the concentration of sodium acetate:
Since the reaction between CH3COOH and NaOH is 1:1, the concentration of sodium acetate formed is equal to the concentration of acetic acid neutralized.
[CH3COONa] = 0.6 × [CH3COOH]0
Calculating the pH of the resulting solution:
The pH of a solution can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where [A-] is the concentration of the conjugate base (sodium acetate) and [HA] is the concentration of the acid (acetic acid).
Substituting the values into the equation, we have:
pH = 4.7 + log ([CH3COONa]/[CH3COOH]remaining)
pH = 4.7 + log (0.6 × [CH3COOH]0 / (0.4 × [CH3COOH]0))
pH = 4.7 + log (0.6 / 0.4)
pH = 4.7 + log (1.5)
pH = 4.7 + 0.18
pH ≈ 4.88
Conclusion:
The pH of the resulting solution after adding NaOH to the CH3COOH solution is approximately 4.88.