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The half-equivalence point of a titration occurs halfway to the equivalence point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.300 moles of a monoprotic weak acid ( Ka=3.6×10−5 M) is titrated with NaOH , what is the pH of the solution at the half-equivalence point?

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ayfat23

Answer:

pH=pKa

pH=4.44

Explanation:

Since the titration occur between a weak acid and a strong base.

then at half -equivalence point, the pH of the solution is equals to the pKa of the weak acid.

Therefore, pH=pKa

Ka of weak acid=3.6×10^−5

To calculate the pKa of the weak acid using the express below;

pKa =- log(Ka)

p​K​a​=​−​l​o​g​(​3.6×10−5)​=​4.44

From the question, the pKa of the solution is at half -equivalence point

Then,

pH=pKa

pH=4.44

adefioyelaoye

The question says that the titration occurred between a weak acid and a

strong base at half-equivalence point. Then we can deduce that the pH of

the solution is equal to the pKa of the weak acid.

pH=pKa

Ka of monoprotic weak acid=3.6×10⁻⁵

The pKa of the monoprotic weak acid will be calculated by :

pKa = - log(Ka)

p​K​a ​=​ −​l​o​g​(​3.6×10⁻⁵)​ =​ 4.44

Since the pKa of the solution is at half -equivalence point

pH=pKa

pH=4.44

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