acid base equilibrium

asked by @jankj2 • over 1 year ago • Chemistry • 5 pts

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.200 M HClO(aq) with 0.200 M KOH(aq). The ionization constant for HClO is 4.0*10^-8

(a) before addition of any KOH (b) after addition of 25.0 mL of KOH (c) after addition of 40.0 mL of KOH (d) after addition of 50.0 mL of KOH (e) after addition of 60.0 mL of KOH

I have two questions: The first is why do I have to set up a rice table for the first question? Can't I simply plug in the answers into pKa=-log(Ka)?

Second one: How do I solve it using the henderson hasslebach equation? do I find the concentration of The initial KOH and then solve? I keep getting wrong answers...

Add comment
1 answer

Hey there. So here are the solutions for this problem. We did the most important ones just so you get an idea of how each one works. For the first part (A), we had no strong base reacting, just a weak acid, so we did a RICE table to show the weak acid reacting and solved for the pH as shown.

For part B, we had some strong base reacting, so we had to use a ICF chart, which must be used with moles. Here, we obtained weak acid and conjugate base, which means we had a buffer. We use the Henderson Hasselbalch equation to determine the pH.

enter image description here

For part D, we show the reaction at the equivalence point. Notice how we only have conjugate base left over, so to solve for the pH, we must use a RICE chart to find the pH. We must find molarity to use in the RICE chart and then simply solved for the pH.

enter image description here

For Part E, we end up having strong base left over, which means that is what we'll use to find our pH. we use the -log of the OH- concentration to find pOH and then we solve for pH. Watch the videos for more help and understanding of this chapter! enter image description here

answered by @sabrina • over 1 year ago