Two 21.0 mL samples, one 0.100 M HCl and the other 0.100 M HF, were titrated with 0.200 M KOH. Answer each of the following questions regarding these two titrations. You may want to reference (Pages 755 - 769) Section 17.4 while completing this problem. Part A What is the volume of added base at the equivalence point for HCl

21 mL of KOH

Explanation:

We have to start with the reaction between HCl and KOH:

Now, using the molarity equation we can find the moles of HCl, (we have to keep in mind that we have to use the volume in L, 21 mL=0.021 L)

Now, we can convert from mol of HCl to mol of KOH using the molar ratio of the balanced reaction. (1 mol HCl= 1 mol KOH)

Finally, we have to calculate the volume of KOH using the molarity equation again:

When we convert this value to mL units we will get:

We will need 21 mL of KOH at the equivalence point.

pie chart because it is in percents ( % )