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.

rosa was looking for patterns to predict the products of chemical reactions. she recorded three similar decomposition reactions in the table.she should record 2licl + 3o2.

in the last row of the table. it is the thermal decomposition of chlorate lithium, hence, she should record these products as: lithium chloride ( 2licl ) and oxygen ( 3o2 )   2 > 2 licl + 3o2.