Question

What volume of 0.117 M HCl is needed to neutralize 28.67ml of 0.137 m KOH?

Answer 1

Approximately 33.6 mL of the 0.117 M HCl solution is needed to neutralize 28.67 mL of 0.137 M KOH.

Step-by-step explanation:

To determine the volume of 0.117 M HCl needed to neutralize 28.67 mL of 0.137 M KOH, we can use the concept of stoichiometry and the balanced chemical equation between HCl and KOH:

HCl + KOH → KCl + H2O

From the balanced equation, we can see that the mole ratio between HCl and KOH is 1:1. This means that 1 mole of HCl reacts with 1 mole of KOH. Therefore, we can calculate the number of moles of KOH present in 28.67 mL of 0.137 M KOH:

Moles of KOH = (0.137 mol/L)(0.02867 L) = 0.0039343 mol KOH

Since the mole ratio is 1:1, the number of moles of HCl required to neutralize the KOH is also 0.0039343 mol. We can calculate the volume of 0.117 M HCl needed using its concentration:

Volume of HCl = (0.0039343 mol)/(0.117 mol/L) = 0.0336 L = 33.6 mL

Therefore, approximately 33.6 mL of the 0.117 M HCl solution is needed to neutralize 28.67 mL of 0.137 M KOH.

Answer 2

To determine the volume of the acid needed in the reaction, we first have to know the reaction involved in the process. Since it is a neutralization reaction, then it should produce a salt and water. The reaction would be:

HCl + KOH = KCl + H2O

We use this reaction to relate the substances. We calculate as follows:

0.137 M KOH (.02867 L solution ) = 3.92779x10^-3 mol KOH
3.92779x10^-3 mol KOH ( 1 mol HCl / 1 mol KOH ) = 3.92779x10^-3 mol HCl

Volume of HCl = 3.92779x10^-3 mol HCl / 0.117 M HCl
= 0.03357 L HCl or 33.57 mL HCl needed