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Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.

Softa [21]

Explanation:

$PbBr_{2}$ will dissociate into ions as follows.

$PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)$

Hence, $K_{sp}$ for this reaction will be as follows.

$K_{sp} = [Pb^{2+}][Br^{-}]^{2}$

We take x as the molar solubility of $PbBr_{2}$ when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

$[Pb^{2+}]$ = $[Pb^{2+}]_{o}$ + x

$[Br^{-}]^{2}$ = $[Br^{-}]_{o}$ + 2x

So, equilibrium solubility expression will be as follows.

$K_{sp}$ = $([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}$

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains $[Br^{-}]_{o}$ = 0.10 and there will be no lead ions. So, $[Pb^{2+}]_{o}$ = 0

So, $[Br^{-}]_{o} + 2x$ will approximately equals to $[Br^{-}]_{o}$ .

Hence, $K_{sp} = x[Br^{-}]^{2}_{o}$

$4.67 \times 10^{-6} = x \times (0.10)^{2}$

x = $4.67 \times 10^{-4}$ M

Thus, we can conclude that molar solubility of $PbBr_{2}$ is $4.67 \times 10^{-4}$ M.

5 0

3 years ago

Read 2 more answers

What accelerates the protons in the LHC?

belka [17]

Answer:

CERN uses a series of particle accelerators to accelerate protons to their end speeds in the LHC. The first accelerator is the linear accelerator Linac 2. At one end of Linac 2, hydrogen from a bottle is passed through an electric field to strip off the electrons, leaving positively charged protons.

Explanation:

3 0

2 years ago

A bowling ball (mass = 7.2 kg, radius = 0.10 m) and a billiard

evablogger [386]

Answer:

Maximum gravitational Force: $F_{gmax} = 1,026*10^{-08} N$

Explanation:

The maximum gravitational force is achieved when the center of gravity are the closer they can be. For the spheres the center of gravity is at the center of it, so the closer this two centers of gravity can be is:

bowling ball radius + billiard ball radius = 0,128 m

The general equation for the magnitude of gravitational force is:

$F_{gmax} = G \frac{M*m}{r_{min}^{2} }$