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

hope this

explanation:

covalent bonds

consist of pairs of electrons shared by two atoms, and bind the atoms to fixed orientation and requre high energies to break them

ionic bonds

when two atoms differ considerably in their electronegativity, then one of the atoms will lose its electron to the other atom. this results in a positively charged ion(cation) and a negatively charged ion (anion)

the answer is the