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 electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.

i think its 5049.

i hope this !