# Amount of Cl- needed to precipitate AgCl given Ksp and [Ag+]

Why is $0.25 \ \mathrm{mL}$ the correct answer in this problem?

I used the Ksp to find the molarity of Chlorine to be $3.47 \times 10^{-2} \ \mathrm{M}$.

Then I used the formula

$$\text{Molarity}_{\text{initial}} \times \text{Volume}_{\text{initial}} = \text{Molarity}_{\text{final}} \times \text{Volume}_{\text{final}}$$

to find the volume of Chlorine, which turned out to be $0.21 \ \mathrm{mL}$ for me. However, my answer does not match the correct answer.

What am I doing wrong?

• Review the calculations – JM97 Sep 16 '16 at 15:44
• 1.8/7.2 = 0.25 exactly – DavePhD Sep 16 '16 at 16:19
• The needed molarity for chlorine that you found is wrong. – MaxW Sep 16 '16 at 19:42

First, solve for [$\ce{Cl-}$] using the expression for K$_{\mathrm sp}$ and the given molarity for $\ce{Ag+}$:

$$K_{\mathrm sp} = 1.8\times 10^{-10} = [\ce{Ag+}][\ce{Cl-}]$$

which rearranges to

$$[\ce{Cl-}] = {1.8\times 10^{-10}\over 7.2\times 10^{-5}} = 2.5\times 10^{-6}\, \mathrm{M}$$

Note that the units for K$_{\mathrm sp}$ are actually $\mathrm{M}^{2}$: I have omitted that from the math above but dimensional analysis tells us that the answer for $[\ce{Cl-}]$ is in units of molarity ($\mathrm{M}$).

Because we have 100 $\mathrm{mL}$ of $\ce{Ag+}$ solution, divide the $[\ce{Cl-}]$ molarity by 10 to get the actual number of moles of $\ce{Cl-}$ needed (here, I multiply by 0.1 to achieve the same result):

$$\mathrm{mol}\,\ce{Cl-} = (2.5\times 10^{-6}\,\mathrm{mol}\cdot\mathrm{L}^{-1}) \times 0.1\,\mathrm{L} = 2.5\times 10^{-7}$$

We are given a solution that is 0.001 $\mathrm{M}$ $\ce{Cl-}$, so we can get the volume needed in liters:

$${2.5\times 10^{-7}\,\mathrm{mol}\,\ce{Cl-}\over 10^{-3}\,\mathrm{M}\,\ce{Cl-}} = 2.5\times 10^{-4}\,\mathrm{L}$$

Finally, convert the above quantity to $\mathrm{mL}$: $$\left({1000\,\mathrm{mL}\over\mathrm{L}}\right)\cdot\,2.5\times 10^{-4}\,\mathrm{L} = 0.25\,\mathrm{mL}$$

• Where did you get the value 0.1 Liters from when trying to convert the molarity of Cl to moles? – Chris B Sep 19 '16 at 14:39
• @ChrisB - That's from the 100 $\mathrm{mL}$ (or 0.1 $\mathrm{L}$) of solution containing the $\ce{Ag+}$, and we need a 1:1 ratio of moles of each species. Since the units of molarity are $\mathrm{mol}\cdot\mathrm{L}^{-1}$, we need to divide by 10 or, as I did, multiply by 0.1. – Todd Minehardt Sep 19 '16 at 15:30