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zubka84 [21]
3 years ago
13

Isopropyl alcohol is mixed with water to produce a 38.0% (v/v) alcohol solution. How many milliliters of each component are pres

ent in 885 mL of this solution? (Assume that volumes are additive.)
Chemistry
1 answer:
Fed [463]3 years ago
5 0
336.3 ml alcohol 548.7 ml of water
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Answer:

Ksp=1.07x10^{-8}

Explanation:

Hello,

In this case, the dissociation reaction is:

PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)

For which the equilibrium expression is:

Ksp=[Pb^{2+}][I^-]^2

Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)

Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M

In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:

[Pb^{2+}]=1.39x10^{-3}M

[I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M

Thereby, the solubility product results:

Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}

Regards.

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