Ag++Clâ’→AgCl(s)âŹâ†“
We need to add over 1 g KCl.
Moles of silver in solution = 1.77â‹…g107.87â‹…gâ‹…molâ’1` = 0.0164â‹…mol
Clearly we need an equivalent quantity of chloride, thus we add a mass of potassium chloride = 0.0164â‹…molĂ—74.55â‹…gâ‹…molâ’1 = ??â‹…g
The 1 equiv will precipitate most of the silver ion. Silver chloride will possess some aqueous solubility. The result is a curdy white precipitate of silver chloride.
