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Savatey [412]
3 years ago
8

Will a precipitate (ppt) form when 300. mL of 2.0 × 10 –5 M AgNO 3 are added to 200. mL of 2.5 × 10 –9 M NaI? Answer yes or no,

and identify the precipitate if there is one
Chemistry
1 answer:
Vsevolod [243]3 years ago
8 0

Answer:

A precipitate will form, AgI

Explanation:

When Ag⁺ and I⁻ ions are in an aqueous media, AgI(s), a precipitate, is produced or not based on its Ksp expression:

Ksp = 8.3x10⁻¹⁷ = [Ag⁺] [I⁻]

<em>Where the concentrations of the ions are the concentrations in equilibrium</em>

For actual concentrations of a solution, you can define Q, <em>reaction quotient, </em>as:

Q = [Ag⁺] [I⁻]

<em>If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form</em>.

Actual concentrations of Ag⁺ and I⁻ are:

[Ag⁺] = [AgNO₃] = 2.0x10⁻⁵ × (300mL / 500.0mL) = 1.2x10⁻⁵M

[I⁻] = [NaI] = 2.5x10⁻⁹ × (200mL / 500.0mL) = 1.0x10⁻⁹M

<em>500.0mL is the volume of the mixture of the solutions</em>

Replacing in Q expression:

Q = [Ag⁺] [I⁻]

Q = [1.2x10⁻⁵M] [1.0x10⁻⁹M]

Q = 1.2x10⁻¹⁴

As Q > Ksp

<h3>A precipitate will form, AgI</h3>

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The explanation:


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At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---&gt; +H2(g) + Br2 (g). If the initial partial pressures o
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Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

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Atoms Hydrogen and Oxygen both have different chemical properties from water H2O and are thus different.

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