If we mix 19.0 mL of a 1.30 ✕ 10−3 M solution of Pb(NO3)2 with 18.6...

If we mix 19.0 mL of a 1.30 ✕ 10⁻³ M solution of Pb(NO₃)₂ with 18.6 mL of a 2.60 ✕ 10⁻³ M solution of NaBr, will a precipitate form?

At this point we would identify the potential precipitate by switching the anions and using the solubility rules to decide if one of the potential products is potentially insoluble. Here the products would be PbBr₂ and NaNO₃. Of these, PbBr₂ is considered "insoluble". The K_sp value for PbBr₂ is 6.60 ✕ 10⁻⁶.

a. Write the Q_sp expression for this potential precipitate.

b. What is the [Br^-] in our mixed solution?

c. What is the value of Q_sp when our two solutions are mixed?

Solutions

Expert Solution

WhenPb(NO3)2 and NaBr are taken , they will dissociate to give individual ions as follows.

Pb(NO₃)₂ Pb²⁺ + 2 NO₃^- ; conc = 1.3x 10^-3 M & volume = 19 ml

Similarly NaBr Na⁺ + Br^- ; conc = 2.6x 10^-3 M & volume = 18.6 ml.

When these two solutions are mixed, total volume = 19 ml + 18.6 ml = 37.6 ml

resulting concentration of individual ions can be calculated as

[Pb2⁺] = 19 x 1.3x 10^-3 / 37.6 = 8.94 x 10^-4M

[NO3^-] = 2 x 19 x 1.3x 10^-3 / 37.6 = 1.79 x 10^-3 M

[Na⁺] = 18.6 x 2.6x 10^-3 / 37.6 = 1.28 x 10-³ M

[Br^-] = 18.6 x 2.6x 10^-3 / 37.6 = 1.28 x 10-³ M

Part A: Since PbBr2 is the potential precipitate, it's Q_sp can be expressed based on the following dissociation process.

PbBr₂ Pb²⁺ + 2 Br^-

Part B: [Br^-] is already calculated above

[Br^-] = 18.6 x 2.6x 10^-3 / 37.6 = 1.28 x 10-³ M

Part C: From the Qsp expression it's value can be calculated as

by substituting the concentration of all the ions

Q_sp = 8.94 x 10^-4M x (1.28 x 10-³ M)²

Q_sp = 14.64 x 10^-10 M³ = 1.464 x 10^-11 M³

Since from the give data and the calculations K_sp > Q_sp We can conclude that PbBr₂ will not precipitate out frrom the solution.

queen_honey_blossom answered 1 year ago

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