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acs-what is a ph at the equivalence point when exactly 25.00 ml of 0.1000 m ch3cooh is titrated with 0.1000 m naoh? ka(ch3cooh)

User Fbmch
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1 Answer

2 votes

Answer:

The pH at the equivalence point is 7.00.

When 25.00 mL of 0.1000 M CH3COOH is titrated with 0.1000 M NaOH, the reaction is:

CH3COOH + NaOH → CH3COO- + H2O

At the equivalence point, the number of moles of CH3COOH is equal to the number of moles of NaOH. This means that the concentration of CH3COO- is equal to the concentration of H+.

The pKa of CH3COOH is 4.75. This means that the pH at the equivalence point is 14 - pKa = 7.00.

Here is the calculation:

pH = -log[H+]

pH = -log[10^(-4.75)]

pH = 7.00

Step-by-step explanation:

User SimpleBeat
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