0.10 M acetic acid is titrated with 0.10 M aqueous ammonia, taking K_a(CH3COOH) = K_b(NH3) = 1.8x10^-5. Select the correct statements.

The pH at the equivalence point is about 7

There is no sharp change of pH near the equivalence point

The pH at half-equivalence is about 4.74

The pH at the equivalence point is about 7

There is no sharp change of pH near the equivalence point

The pH at half-equivalence is about 4.74

Phenolphthalein (range 8.3-10.0) is a suitable indicator

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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ia054

0.10 M acetic acid is titrated with 0.10 M aqueous ammonia, taking $K_{a} (CH X_{3} COOH) = K_{b} (NH X_{3}) = 1.8 \times 1 0^{- 5}$ . Select the correct statements.

AThe pH at the equivalence point is about 7
BThere is no sharp change of pH near the equivalence point
CThe pH at half-equivalence is about 4.74
DPhenolphthalein (range 8.3-10.0) is a suitable indicator

Answer: A, B, C

The product

CH X_{3} COONH X_{4}

has

K_{a} = K_{b}

, so pH

\approx 7

. Both ions hydrolyse and the solution is self-buffered, so the pH rises gradually near equivalence and no indicator gives a sharp end point; phenolphthalein is not suitable. At half-equivalence acetic acid and acetate are equal, pH

\approx p K_{a} = 4.74

. Answer A, B, C.

Practice Ionic Equilibrium II More on Acid-base titration curves and indicators

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0.10 M acetic acid is titrated with 0.10 M aqueous ammonia, taking Ka​(CHX3​COOH)=Kb​(NHX3​)=1.8×10−5. Select the correct statements.

Answer: A, B, C

More questions on Acid-base titration curves and indicators

0.10 M acetic acid is titrated with 0.10 M aqueous ammonia, taking $K_{a} (CH X_{3} COOH) = K_{b} (NH X_{3}) = 1.8 \times 1 0^{- 5}$ . Select the correct statements.