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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.

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

                     PbBr₂ ⇌ Pb²⁺  +  Br⁻

I/mol·L⁻¹:                    0.0140  0.0024

[tex]Q_{\text{sp}}[/tex] = [Pb²⁺][Br⁻]²

[Pb²⁺] = 0.0140 mol·L⁻¹

[Br⁻] = 0.0024 mol·L⁻¹

[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^⁻7.

What is concentration?

Concentration refers to the amount of substance present in solution. We have a solution of Pb(NO₃)₂ and NaBr. The two solutions react to produce an insoluble product which is  PbBr₂.

The equilibrium is:

PbBr₂(s) ⇄ Pb²⁺(aq) + 2Br⁻(aq)

Note that :

Q = [Pb²⁺] [Br⁻]²

Also:

[Pb²⁺] = 0.0024M

[Br⁻] = 0.0140M

Hence;

Q = [Pb²⁺] [Br⁻]²

Q = [0.0024M] [0.0140M]²

Q = 4.7x10^⁻7

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

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