# A solution contains 1.77 g of dissolved silver. How many moles of potassium chloride must be added to the solution to completely precipitate all of the silver? What mass of potassium chloride must be added?

May 22, 2016

$A {g}^{+} + C {l}^{-} \rightarrow A g C l \left(s\right) \downarrow$

We need to add over $1$ $g$ $\text{KCl}$.

#### Explanation:

$\text{Moles of silver in solution}$ $=$ (1.77*g)/(107.87*g*mol^-1`) $=$ $0.0164 \cdot m o l$

Clearly we need an equivalent quantity of chloride, thus we add a mass of potassium chloride:

$= 0.0164 \cdot m o l \times 74.55 \cdot g \cdot m o {l}^{-} 1$ $=$ ??*g

Of course addition of the 1 equiv will precipitate most of the silver ion. Silver chloride will possess some (low) aqueous solubility, according to its solubility product. The result is a curdy white precipitate of silver chloride, that will be very hard to handle.