<p>A 0.052 M solution of benzoic acid (equilibrium constant Ka = 6.3 * 10^-5) is titrated with a strong base. What is the [H+] ofthe solution one-half way to the equivalence point?</p>
<p>Answer: 6.3 * 10^-5</p>
<p>My teacher gave me the answer and asked us to find out procedures by ourself; but I have no clue how to start.</p>
<p>Please help!!!!!! Anybody?</p>
<p>The Henderson-Hasselbach equation is: pH = pKa + log ([base]/[acid]).</p>
<p>And when you titrate, at the half-equivalence point, pH = pKa. </p>
<p>Easier than it seems, eh?</p>
<p>thanks, but what does "half-equivalence point" mean?</p>
<p>Is it when the acid lose half of its concentration?</p>
<p>"the solution one-half way to the equivalence point".</p>
<p>So if you draw the titration curve: put the mL of base used on the x-axis and pH on the y-axis. If, theoretically, it took 40 mL to titrate the acid, then the half-equivalence point is at 20 mL. </p>
<p>Simple definition? I suppose ... the point at which half the volume of base is added before reaching equivalence.</p>