How many hydroxide ions are needed to completely neutralize 1.0 liter of 0.50 M HCI?

Question

Cameron Andrews · Accepted Answer

#"Half an Avocado number...."#

We deal with the equation of stoichiometry.

#HCl(aq) + NaOH(aq) rarr NaCl(aq) + H_2O(l)#

We have #1.0*Lxx0.50*mol*L^-1-=0.50*mol# with respect to (formal!) hydrogen chloride molecules. Stoichiolmetric equivalence requires an equal number of hydroxides....i.e. #1/2*molxxunderbrace(6.022xx10^23*mol^-1)_"Avogadro's number"# . Claro?

Naomi Aronson · Answer

undefined To completely neutralize 1.0 liter of 0.50 M HCl, we need to determine the ratio of hydroxide ions (OH-) to hydrogen ions (H+) in the reaction. Since HCl is a strong acid and completely dissociates in water, it produces one hydrogen ion for every molecule of HCl. Therefore, the molarity of H+ ions in the solution is equal to the molarity of the HCl solution, which is 0.50 M.

To neutralize H+ ions, we need an equal number of OH- ions. The neutralization reaction between H+ and OH- ions forms water (H2O).

So, for every 1 mole of H+ ions from HCl, we need 1 mole of OH- ions to neutralize them.

Given that the volume of the solution is 1.0 liter and the molarity of H+ ions is 0.50 M, the number of moles of H+ ions is:

Moles of H+ ions = Molarity * Volume = 0.50 mol/L * 1.0 L = 0.50 moles

Since the ratio of H+ ions to OH- ions is 1:1, we need the same number of moles of OH- ions. Therefore, we need 0.50 moles of OH- ions to completely neutralize 1.0 liter of 0.50 M HCl.

However, since the question asks for the number of hydroxide ions, we need to consider Avogadro's number. 1 mole of hydroxide ions (OH-) contains 6.022 x 10^23 ions.

So, the number of hydroxide ions needed is:

0.50 moles * 6.022 x 10^23 ions/mole ≈ 3.011 x 10^23 hydroxide ions.

Therefore, approximately 3.011 x 10^23 hydroxide ions are needed to completely neutralize 1.0 liter of 0.50 M HCl.