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A solution containing lead(ii) nitrate is mixed with one containing sodium bromide to form a solution that is 0.0140 m in pb(no3)2 and 0.0024 m in nabr. what is the value of q for the insoluble product? express the reaction quotient to three significant figures.

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Respuesta :

znk

Answer:

[tex]Q_{\text{sp}} = 8.06 \times 10^{-8}[/tex]

Explanation:

Since all sodium salts and all nitrates are soluble, the insoluble product must be lead(II) bromide.

The equation for the equilibrium is

Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā PbBrā‚‚ ā‡Œ PbĀ²āŗ Ā + Ā Brā»

I/molĀ·Lā»Ā¹: Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā 0.0140 Ā 0.0024

[tex]Q_{\text{sp}}[/tex] = [PbĀ²āŗ][Brā»]Ā²

[PbĀ²āŗ] = 0.0140 molĀ·Lā»Ā¹

[Brā»] = 0.0024 molĀ·Lā»Ā¹

[tex]Q_{\text{sp}} = 0.0140 \times 0.0024^{2}[/tex]

[tex]Q_{\text{sp}} = 8.06 \times 10^{-8}[/tex]

pstnonsonjoku

The reaction quocient is 4.7x10^ā»7.

What is concentration?

Concentration refers to the amount of substance present in solution. We have a solution of Pb(NOā‚ƒ)ā‚‚ and NaBr. The two solutions react to produce an insoluble product which is Ā PbBrā‚‚.

The equilibrium is:

PbBrā‚‚(s) ā‡„ PbĀ²āŗ(aq) + 2Brā»(aq)

Note that :

Q = [PbĀ²āŗ] [Brā»]Ā²

Also:

[PbĀ²āŗ] = 0.0024M

[Brā»] = 0.0140M

Hence;

Q = [PbĀ²āŗ] [Brā»]Ā²

Q = [0.0024M] [0.0140M]Ā²

Q = 4.7x10^ā»7

Learn more about concentration: https://brainly.com/question/10468518