Question

Suppose our unknown acid was acetic acid, CH3COOH (Ka = 1.8 x 10-5) and that your initial aqueous solution contained 0.100 mol of acetic acid in total volume of 100 mL.

To this acetic acid solution you begin adding a 1.0 mol/L NaOH solution. Let's see what happens after adding different amounts of NaOH.

Suppose you have added a total of 20.0 mL of NaOH solution since the beginning of the titration. How many mol of acetate ions (CH3COO–) are present in the reaction mixture at this point?

Answer 1

Answer on Question #63492 - Chemistry | General Chemistry

Suppose our unknown acid was acetic acid, $\mathrm{CH_3COOH}$ ($\mathrm{Ka} = 1.8\cdot 10^{-5}$) and that your initial aqueous solution contained $0.100\ \mathrm{mol}$ of acetic acid in total volume of $100\ \mathrm{mL}$.

To this acetic acid solution you begin adding a $1.0\ \mathrm{mol/L}\ \mathrm{NaOH}$ solution. Let's see what happens after adding different amounts of $\mathrm{NaOH}$.

Suppose you have added a total of $20.0\ \mathrm{mL}$ of NaOH solution since the beginning of the titration. How many mol of acetate ions ($\mathrm{CH_3COO^-}$) are present in the reaction mixture at this point?

Solution

In dealing with weak acids, there will be an incomplete dissociation in water.

$$\mathrm{CH_3COOH} \leftrightarrows \mathrm{CH_3COO^-} + \mathrm{H^+}$$

So your initial aq solution has $0.100\ \mathrm{mol}$ of acetic acid and,

$$n = 1.0\ \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.0200\ \mathrm{L} = 0.020\ \mathrm{mol}\ \text{of}\ \mathrm{NaOH}.$$

The added base will neutralize some of the acid, and produce the conjugate base, $\mathrm{CH_3COO^-}$ and $\mathrm{H_2O}$. At this point, we have neutralized 0.020 moles of the acid and produced 0.020 moles of $\mathrm{CH_3COO^-}$.

Answer

0.020 mol of acetate ions ($\mathrm{CH_3COO^-}$) are present in the reaction mixture.

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