Question

The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.580 moles of a monoprotic weak acid (Ka = 4.5 × 10-5) is titrated with NaOH, what is the pH of the solution at the half-equivalence point?

Answer 1

you can figure this out by using the Henderson-Hasselbalch equation:
\(\color{red}{pH=pKa+log\dfrac{[A-]}{{[HA]}}}\)

the Ka is given to you can find the pKa: \(pKa=-log(K_a)\)

the amount of acid (in moles) is given.
At the half equivalence point half of it is in the unprotonated form and half in the protonated form. So HA=0.580/2 moles and \(A^-\)=0.580/2 moles

Assuming one liter \([HA]=0.580/2 =[A^-]\)

pH is what you're looking for in the question, plugging your values in:

\(pH=-log(4.5*10^{-5})+log\dfrac{0.580/2 }{{[0.580/2 ]}}\)
\(pH=4.3467 +log(1)\)
\(pH=4.3467+ 0\)
\(pH=4.35\)

\(\large\sf\color{green}{So\;at\;the \;half-equivalence\; point \;pH=pKa, \small\;this\; is\; applies\;to\;any\; weak\; acid.}\)