A 50.0 mL solution of Ba(OH)2 is combined with a 150 mL solution of 0.20 M HCl. If the resulting solution has a hydroxide ion concentration of 0.12 M, what was the concentration of Ba(OH)2 in the original solution?

0.54M of Ba(OH)2

Explanation:

When Ba(OH)2 reacts with HCl, BaCl2 and H2O are produced as follows:

Ba(OH)2 + 2HCl → BaCl2 + 2H2O

The remanent hydroxide ion is because not all Ba(OH)2 reacts. Thus, we need to find moles of Ba(OH)2 that doesn't react and moles of Ba(OH)2 that reacts. The ratio between total moles and volume of the Ba(OH)2 solution = 0.050L is the molarity of the original solution

Moles of Ba(OH)2 that doesn't react:

50mL + 150mL = 0.200L * (0.12 mol OH- / L) = 0.024 moles OH-

2 moles of OH- are in 1 mole of Ba(OH)2:

0.024 moles OH- * (1mol Ba(OH)2 / 2 mol OH-) = 0.012 moles Ba(OH)2

Moles of Ba(OH)2 that react:

0.150L * (0.20mol / L) = 0.030 moles HCl

2 moles of HCl react per mole of Ba(OH)2:

0.030 moles HCl * (1mol Ba(OH)2 / 2 mol HCl) = 0.015 moles Ba(OH)2

Total moles:

0.012mol + 0.015mol = 0.027mol Ba(OH)2 in 50mL

0.027mol Ba(OH)2 / 0.0500L =

0.54M of Ba(OH)2

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.

pie chart because it is in percents ( % )