You could add HCl(aq) to the solution to precipitate out AgCl(s). What volume of a 0.190 M HCl(aq) solution is needed to precipitate the silver ions from 12.0 mL of a 0.160 M AgNO3 solution? Express your answer with the appropriate units.

10.1mL

Explanation:

To solve this problem, let's generate the equation:

HCl + AgNO3 —> AgCl + HNO3

Next, we find the amount of AgNO3 in 12mL of the solution.

12mL = 0.012L

If 0.16mol of AgNO3 dissolves in 1L of solution,

Therefore Xmol will dissolve in 0.012L i.e

Xmol of AgNO3 = 0.16x0.012 = 0.00192mol

From the equation, 1mole of AgNO3 required 1mole HCl.

This implies that 0.00192mol of AgNO3 will also require 0.00192mol of HCl.

From this result, we can calculate the Volume of HCl that is needed for the reaction. This is done by:

Molarity = mole /Volume

Molarity = 0.19M

Mole = 0.00192mol

Volume =?

Volume = mole / Molarity

Volume = 0.00192 / 0.19

Volume = 0.0101L = 10.1mL

Therefore the volume of HCl needed is 10.1mL

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6.4 cm (max) - 25 cm (min)