Question

Will a precipitate form when 20.0 mL of Pb(NO3)2 with a concentration of 2.50 x 10^-4 mol/L is mixed with 45.0 mL of CaCl2 with a concentration of 1.25 x 10^-3 mol/L?

Answer 1

Question #57947, Chemistry, Other

Will a precipitate form when $20.0\,\mathrm{mL}$ of $\mathrm{Pb(NO_3)_2}$ with a concentration of $2.50 \times 10^{-4}\,\mathrm{mol/L}$ is mixed with $45.0\,\mathrm{mL}$ of $\mathrm{CaCl_2}$ with a concentration of $1.25 \times 10^{-3}\,\mathrm{mol/L}$?

Answer:

The expected precipitate is:

$$\mathrm{Pb(NO_3)_2} + \mathrm{CaCl_2} = \mathrm{PbCl_2} \downarrow + \mathrm{Ca(NO_3)_2}$$

The precipitate will form when the concentration of $\mathrm{PbCl_2}$ will exceed the $K_{\mathrm{sp}}$.

$K_{\mathrm{sp}}$ for $\mathrm{PbCl_2}$ is $1.7 \times 10^{-5}$

$$K_{\mathrm{sp}} = [\mathrm{Pb^{2+}}][\mathrm{Cl^-}]^2$$

For \mathrm{Pb(NO_3)_2: 0.00025·20.0=C_2·1000\,\mathrm{ml}$

$$\mathrm{C_2(Pb(NO_3)_2)} = 0.000005\,\mathrm{M}$$

For $\mathrm{CaCl_2}$: 0.00125·45.0=C_2·1000\,\mathrm{ml}

$$\mathrm{C_2(CaCl_2)} = 0.000056\,\mathrm{M}$$ $$Q = 0.000005 \cdot 0.000056^2 = 1.57 \cdot 10^{-14}$$

As long as the obtained value is less than $K_{\mathrm{sp}}$ for $\mathrm{PbCl_2}$, the precipitate will not form.

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