Chemistry, 20.07.2019 15:30 barn01

Asolution is prepared by mixing 50.0 ml of 0.50 m cu(no3)2 with 50.0 ml of 0.50 m co(no3)2. sodium hydroxide is slowly added to the mixture. which precipitates first? [ksp(cu(oh)2) = 2.2 × 10–20, ksp(co(oh)2) = 1.3 × 10–15]

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In the mixture of 50.0 mL of 0.50 M Cu(NO₃)₂ and 50.0 mL of 0.50 M Co(NO₃)₂, after the addition of NaOH the Cu(OH)₂ will precipitate first.

To find which precipitates first, we need to calculate the concentration of OH⁻ in the solution.

1. Cu(OH)₂

Cu(OH)₂(s) → Cu²⁺(aq) + 2OH⁻(aq) (1)

The product of solubility constant of the above reaction is:

$Ksp_{Cu(OH)_{2}} = [Cu^{2+}][OH^{-}]^{2}$ (2)

The concentration of Cu²⁺ in the mixture can be calculated as follows:

$n_{Cu(NO_{3})_{2}} = C_{Cu(NO_{3})_{2}}_{i}*V_{Cu(NO_{3})_{2}}$

Where:

n: is the number of moles

C: is the concentration

V: is the volume

The number of moles of Cu(NO₃)₂ is:

$n_{Cu(NO_{3})_{2}} = 0.50 mol/L*0.050 L = 0.025 \:moles$

When the Cu(NO₃)₂ is mixed with 50.0 mL of Co(NO₃)₂, the final concentration of Cu(NO₃)₂ is:

$C_{Cu(NO_{3})_{2}}_{f} = \frac{n_{Cu(NO_{3})_{2}}}{V_{t}} = \frac{0.025\: moles}{(0.050 + 0.050) L} = 0.25 mol/L$

Now, we can find the value of [OH⁻] (eq 2)

$[OH^{-}] = \sqrt{\frac{Ksp_{Cu(OH)_{2}}}{[Cu^{2+}]}} = \sqrt{\frac{2.2 \cdot 10^{-20}}{0.25}} = 2.97 \cdot 10^{-10} mol/L$

2. Co(OH)₂

Co(OH)₂(s) → Co²⁺(aq) + 2OH⁻(aq) (3)

The product of solubility constant of reaction (3) is:

$Ksp_{Co(OH)_{2}} = [Co^{2+}][OH^{-}]^{2}$ (4)

The concentration of Co²⁺ after the mixture with Cu(NO₃)₂ is:

$n_{Co(NO_{3})_{2}} = 0.50 mol/L*0.050 L = 0.025 \:moles$

$C_{Co(NO_{3})_{2}}_{f} = \frac{n_{Co(NO_{3})_{2}}}{V_{t}} = \frac{0.025\: moles}{(0.050 + 0.050) L} = 0.25 mol/L$

So, the concentration of OH⁻ when it reacts with Co(NO₃)₂ is (eq 4):

$[OH^{-}] = \sqrt{\frac{Ksp_{Co(OH)_{2}}}{[Co^{2+}]}} = \sqrt{\frac{1.3 \cdot 10^{-15}}{0.25}} = 7.21 \cdot 10^{-8} mol/L$

Since the concentration of OH⁻ needed to form Cu(OH)₂ is lower than the concentration needed to form Co(OH)₂, the Cu(OH)₂ will precipitate first.

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Answer from: love7973

When we dilute both molarities of Cu(NO3) and CO(NO3)2 to half

to become 0.25 M Cu(NO3) & 0.25 M CO(NO3)2

and when the value of Ksp CO(OH)2 = 1.6 x 10^-15

and the value of Ksp Cu(OH)2 = 2.2 x 10^-20

1) so, for CO(OH)2 Ksp expression = [CO+2][OH-]^2

by substitution:

1.6 x 10^-15 = 0.25 m * [OH-]^2

∴ [OH] = 8 x 10^-8

∴POH = -㏒[OH-]

= -㏒ 8 x 10^-8

= 7.1
∴ PH= 14 - POH

PH = 14 - 7.1

= 6.9

2) and for Cu(OH)^2

Ksp expression = [Cu+2][OH-]^2

by substitution:

2.2 x 10^-20 = 0.25 * [OH-]^2

∴[OH-] = 2.97 x 10^-10

POH = -㏒[OH]

= -㏒(2.97 x 10^-10)

= 9.5

∴PH = 14 - POH

= 4.5

∴ Cu(OH)2 will precipitate first because it needs less [OH-] and PH values than CO(OH)2 to reach the Ksp value and reach equilibrium.

So you have to add more [OH-] than 2.97 x 10^-10 to precipitate Cu(OH)2

and you have to add more [OH-] than 8 x 10^-8 to precipitate CO(OH)2
Asolution is prepared by mixing 50.0 ml of 0.50 m cu(no3)2 with 50.0 ml of 0.50 m co(no3)2. sodium h

Asolution is prepared by mixing 50.0 ml of 0.50 m cu(no3)2 with 50.0 ml of 0.50 m co(no3)2. sodium h

Answer from: Quest

Ur e n d is not real

Answer from: Quest

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.

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