Topic: Ksp problem (Read 1700 times)

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

K_{s p} = [{Pb}^{2 +}] \cdot [{Cl}^{-}]^{2}

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

S = \sqrt[3]{\frac{K_{s p}}{4}} = 1.59 \cdot 10^{- 2}

$\displaystyle S = \sqrt[3]{\dfrac{K_\mathrm{sp}}{4}} = 1.59 \cdot 10^{-2}$

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

Borek · « **Reply #1 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.

Mimic · « **Reply #2 on:** February 10, 2017, 06:49:20 AM »

K_{s p} = [{Pb}^{2 +}] \cdot [{Cl}^{-}]^{2}

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

1.62 \cdot 10^{- 5} = [{Pb}^{2 +}] \cdot [{Cl}^{-}]^{2}

$1.62 \cdot 10^{-5} = [\ce{Pb^2+}] \cdot [\ce{Cl-}]^2$

1.62 \cdot 10^{- 5} = 0.3 M \cdot [{Cl}^{-}]^{2}

$1.62 \cdot 10^{-5} = 0.3M \cdot [\ce{Cl-}]^2$

[{Cl}^{-}]^{2} = \frac{1.62 \cdot 10^{- 5}}{0.3 M}

$[\ce{Cl-}]^2 = \dfrac{1.62 \cdot 10^{-5}}{0.3M}$

[{Cl}^{-}] = \sqrt{\frac{1.62 \cdot 10^{- 5}}{0.3 M}} = 7.35 \cdot 10^{- 3} M

$[\ce{Cl-}] = \sqrt{ \dfrac{1.62 \cdot 10^{-5}}{0.3M} } = 7.35 \cdot 10^{-3} M$

It's right?

AWK · « **Reply #3 on:** February 10, 2017, 07:31:25 AM »

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

Topic: Ksp problem (Read 1700 times)

Mimic

Ksp problem

Borek

Re: Ksp problem

Mimic

Re: Ksp problem

AWK

Re: Ksp problem

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