Chemistry 1151 lab

Question

How many grams of KOH are needed to neutralize 10.2 mL of 0.14 M HCl in stomach acid?

Answer 1

This problem involves an Acid Base neutralization reaction. This class of reaction typically yields some type of salt, as well as water. We can see this from the balanced reaction below, where KCl is the salt produced:

KOH + HCl --> KCl + H2O

The problems tells us that we initially have 10.2mL (0.0102 L) of 0.14M HCl. The ratio of H to Cl in the HCl chemical formula is 1:1. Because of this, when the HCl dissociates in solution to H+ and Cl-, we know that for every mole of HCl that dissociates, it will yield exactly 1 mole of H+ and 1 mole of Cl-.

With this in mind, if we multiply the volume of acid (in L) by our concentration (moleHCl/L), the product will yield the number of moles of HCl present in the solution, and subsequently the number of moles of H+ ions that will need to be "neutralized" by our KOH base.

So, (0.0102 L)*(0.14M) = 0.001428 mole of HCl in solution = 0.001428 mole of H+ ions in solution

We can see from the stoichiometry of the balanced reaction above that for every mole of HCl consumed, 1 mole of KOH is also consumed. This means that the 0.001428 moles of H+ ions in solution will require exactly 0.001428 moles of OH- ions to be fully neutralized via its conversion into water. (1H+ +1OH- --> 1H2O)

All we have to do now is to use this number of moles of KOH along with its molar mass to find the total mass of KOH required for this neutralization. Similar to the HCl, the ratio of K to OH in the KOH chemical formula is 1:1. Because of this, when the KOH dissociates in solution into K+ and OH-, we know that for every mole of KOH that dissociates, it will yield exactly 1 mole of OH- and 1 mole of K+.

With this in mind, we can say that 0.001428 moles of KOH are required for this neutralization reaction.

The molar mass of KOH is 56.1056 g/mole. If we multiply the number of moles of KOH by its molar mass, the product will yield the total mass of KOH required.

So, (0.001428 mole KOH)*(56.1056 g/mole) = 0.08012 g of KOH.