# Estimate the concentration of HCl and volume of 6 M HCl required to complete precipitate each of the given salts

In our chemistry lab we performed the following experiment below.

Given that $K_\mathrm{sp}(\ce{AgCl}) = 8.2 \times 10^{-11}$ and $K_{\mathrm{sp}}(\ce{PbCl_{2}}) = 1.7 \times 10^{-5}$

Procedure: $0.2\,\mathrm{mL}$ each of $\mathrm{0.05\: M \: \ce{Ag+}}$ ions and $\mathrm{ 0.05\: M \: \ce{Pb^{2+}}}$ ions were added into a centrifuge tube. The solution was diluted to $1\,\mathrm{mL}$ $6\,\mathrm{M}$ $\ce{HCl}$ was added to the solution till it completely precipitates. The quantity of $\ce{HCl}$ required was recorded. The precipitate and supernatant were separated and kept for subsequent experiments.

Observations: Precipitate formation was observed almost instantly. It took 6 drops of $\ce{HCl}$ to completely precipitate the contents.

Estimate the concentration of $\ce{HCl}$ and volume of $6\,\mathrm{M}$ $\ce{HCl}$ required to complete precipitate each of the given salts.

Here is my approach:

\begin{align} c_{1}V_{1} &= c_{2}V_{2} \\ \text{Let } c_{1}=0.05\:\mathrm{M}, V_{1}&=0.2\:\mathrm{mL}, \text{and} \ V_{2}=1\:\mathrm{mL}\\ [\ce{Pb^2+}] &= 0.2 \cdot 0.05\ \mathrm{M} \\ [\ce{Ag+}] &= 0.2 \cdot 0.05\ \mathrm{M}\\ K_{\mathrm{sp}}\:(\ce{AgCl}) &= [\ce{Ag+}] [\ce{Cl-}] \\ 8.2 \times 10^{-11} &= 0.2 \cdot 0.05\ \mathrm{M} \cdot s_{1} \\ s_{1} &= 8.2 \times 10^{-9}\ \mathrm{M}\\ K_{\mathrm{sp}}\:(\ce{PbCl_{2}}) &= [\ce{Pb^{2+}}] [\ce{Cl-}]^{2}\\ 1.7 \times 10^{-5} &= 0.2 \cdot 0.05\ \mathrm{M} \cdot (s_{2})^2\\ s_{2} &= 0.041\ \mathrm{M} \end{align}

Thus to precipitate $\ce{AgCl}$ and $\ce{PbCl_{2}}$ completely, the concentration of chloride ions in solution must exceed $8.2 \times 10^{-9}$ M and $0.041$ M respectively).

I am not sure of my computations up till this point, and I need some help on how to proceed further.

EDIT: My calculations tell me that volume of $6\,\mathrm{M}\,\ce{HCl}$ necessary is $0.00683\,\mathrm{mL}$. In the lab we have estimated the volume of one drop of our pipette as $0.037\,\mathrm{mL}$, based on this I calculated that the number of drops of hydrochloric acid necessary to complete precipitation is $0.18$ drops. But in the actual experiment I performed 6 drops of $\ce{HCl}$ were dispensed and based on the computations we did, everything should have precipitated. But after centrifugation and separation the supernatant still had lead ions which were precipitated as lead chromate. Why is this so? (This is the main reason why I suspected my calculations and turned to this website for help).

EDIT2: I now see the flaw in my computations. what I have calculated is in fact that amount of HCl necessary to start precipitation. Thanks to everyone for their help.

What you got are two different concentrations required to start to precipitate all the cations in your solution. However, you are asked to completely precipitate the cations. Well — that is a nigh impossible task unless you define a concentration threshold which you want to reach. Your threshold could be $1\,\mathrm{mM}$, $1\,\mathrm{µM}$, $1\,\mathrm{nM}$ or whichever you see fit. Use this desired concentration to recalculate the concentration of chloride ions required for both cations. Then continue on.
I recalculated the value in the spoiler tag for $1\,\mathrm{µM}$ residual cation concentration.
$4.12\,\mathrm{M}$ chloride, equivalent to $0.687\,\mathrm{mL}$ or 18 drops.
Backwards calculation tells me, that you had a residual lead concentration of ca. $9.58\,\mathrm{µM}$, explaining your second result.