Answer:
Step-by-step explanation:
First particle drawing:
Ba(OH)2 (aq) + H2SO4 (aq)
o o
/ \ / \
Ba Ba H S(=O)2-OH
\ / \ /
o o
Reactants together in the container prior to any reaction taking place. Ba(OH)2 (aq) and H2SO4 (aq) are present in the container. The Ba(OH)2 particles are shown as two Ba atoms connected to each other by a chemical bond and surrounded by hydroxide (OH-) ions, represented by circles. The H2SO4 particles are shown as two H atoms connected to a S atom and two O atoms connected to the S atom, each with a single bond, and an -OH group connected to one of the O atoms, represented by circles.
Second particle drawing:
o o
/ \ / \
Ba S(=O)2-OH
/ \ / \
OH OH OH OH
Species present at the equivalence point. The reaction is complete, and all of the Ba(OH)2 and H2SO4 have reacted to form BaSO4 (barium sulfate) and H2O (water). The particles of BaSO4 are shown as a Ba atom connected to an S atom and four O atoms, each with a single bond, represented by circles, and the particles of H2O are shown as two H atoms connected to an O atom, represented by circles.
Third particle drawing:
o o
/ \ / \
Ba S(=O)2-OH
/ \ / \
OH OH OH OH
/
H+
Species present after the equivalence point while sulfuric acid is still added. All the barium hydroxide has been consumed, but some sulfuric acid is still present. The H+ ions from the sulfuric acid combine with the OH- ions in the solution to form water (H2O), represented by circles. The BaSO4 particles are still present in the solution.