Predicting precipitation

Step 1 — Identify the possible precipitate

Mixing $\text{AgNO}_3$ and $\text{NaCl}$ could give AgCl or $\text{NaNO}_3$.

Sodium nitrate is very soluble, so the only candidate is silver chloride, and the equilibrium to consider is:

$$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq) \qquad K_s = [\text{Ag}^+][\text{Cl}^-]$$

Step 2 — Find the moles of each ion before mixing

Each salt supplies one of the relevant ion per formula unit.

$$n(\text{Ag}^+) = cV = 0.0020 \times 0.0250 = 5.00 \times 10^{-5} \text{ mol}$$

$$n(\text{Cl}^-) = cV = 0.0015 \times 0.0350 = 5.25 \times 10^{-5} \text{ mol}$$

Step 3 — Find the total volume

This is the step that must not be skipped.

$$V_{\text{total}} = 25.0 + 35.0 = 60.0 \text{ mL} = 0.0600 \text{ L}$$

Step 4 — Find the concentration of each ion after mixing

$$[\text{Ag}^+] = \frac{5.00 \times 10^{-5}}{0.0600} = 8.33 \times 10^{-4} \text{ mol L}^{-1}$$

$$[\text{Cl}^-] = \frac{5.25 \times 10^{-5}}{0.0600} = 8.75 \times 10^{-4} \text{ mol L}^{-1}$$

Both are lower than in the original solutions, because each has been diluted by the other.

Step 5 — Calculate $Q$

$$Q = [\text{Ag}^+][\text{Cl}^-] = (8.33 \times 10^{-4})(8.75 \times 10^{-4})$$

$$Q = 7.29 \times 10^{-7}$$

Step 6 — Compare and conclude

$$Q = 7.29 \times 10^{-7} \qquad K_s = 1.8 \times 10^{-10}$$

$$Q > K_s \quad \text{by a factor of about } 4000$$