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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Mimic on February 10, 2017, 06:04:36 AM

Title: Ksp problem
Post by: Mimic on February 10, 2017, 06:04:36 AM

How many mL of HCl 0.3M I must add to 5 mL di Pb(NO₃)₂ 0.3M to start the precipitation of PbCl₂, knowing that its Ksp is 1.62·10^-5?

From Ksp formula

[tex]K_\mathrm{sp} = [\ce{Pb^2+}] \cdot [\ce{Cl-}]^2[/tex]
[tex]\displaystyle S = \sqrt[3]{\dfrac{K_\mathrm{sp}}{4}} = 1.59 \cdot 10^{-2}[/tex]

S is the concentration of ions in mol/L above which begins the formation of the precipitate. Knowing this, what should I do?

Title: Re: Ksp problem
Post by: Borek on February 10, 2017, 06:37:26 AM

It is not about solubility, but about concentrations of individual ions. Try to express them (from dilutions) as a function of the volume of acid added.

Title: Re: Ksp problem
Post by: Mimic on February 10, 2017, 06:49:20 AM

[tex]K_\mathrm{sp} = [\ce{Pb^2+}] \cdot [\ce{Cl-}]^2[/tex]
[tex]1.62 \cdot 10^{-5} = [\ce{Pb^2+}] \cdot [\ce{Cl-}]^2[/tex]
[tex]1.62 \cdot 10^{-5} = 0.3M \cdot [\ce{Cl-}]^2[/tex]
[tex][\ce{Cl-}]^2 = \dfrac{1.62 \cdot 10^{-5}}{0.3M} [/tex]
[tex][\ce{Cl-}] = \sqrt{ \dfrac{1.62 \cdot 10^{-5}}{0.3M} } = 7.35 \cdot 10^{-3} M[/tex]

It's right?

Title: Re: Ksp problem
Post by: AWK on February 10, 2017, 07:31:25 AM

When you add acid both concentrations: Pb²⁺ and Cl^- are changing.