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?

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)

pKa=−log(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

a galvanic cell( aka voltaic cell) and electrolytic cell

explanation:

stan bts (kpop group)

follow my insta yoongi_jams

subscribe to my yt yoongi jmz