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4 years ago

The Ksp for Cu(OH)2 is 4.8 × 10-20. Determine the molar solubility of Cu(OH)2 in a buffer solution with a pH of 10.1.

Chemistry

1 answer:

Papessa [141]4 years ago

3 0

Answer:

E) 3.0x10⁻¹²

Explanation:

Ksp of Cu(OH)₂ is defined as:

Cu(OH)₂(s) ⇄ Cu²⁺(aq) + 2 OH⁻(aq)

Ksp = [Cu²⁺] [OH⁻]²

When pH is 10.1, [OH⁻] is:

pOH = 14 - pH

pOH = 3.9

pOH = -log [OH⁻]

[OH⁻] = 1.26x10⁻⁴M

Replacing in ksp formula:

4.8x10⁻²⁰= [Cu²⁺] [1.26x10⁻⁴]²

3.0x10⁻¹² = [Cu²⁺]

That means the maximum amount of Cu²⁺ that can be in solution is 3.0x10⁻¹²M, thus, molar solubility of Cu(OH)₂ is

E) 3.0x10⁻¹²

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How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

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3 years ago

An example of a substance that can be decomposed by a chemical change is:

Savatey [412]

The answer is iron..

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4 years ago

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What is the volume of water in 150ml of the 35% of sucrose with a specific gravity of 1.115?

anzhelika [568]

Answer:

The volume of the water is 108.71 mL

Explanation:

Step 1: Data given

Volume of water =150 mL = 0.150 L

concentration of sucrose solution 35 % w/w this means in 100 grams of water we have 35 grams of sucrose

specific gravity =1.115

Step 2: Calculate the density of the solution

Density = specific gravity * density of water

Density of solution = 1.115 * 1g/ mL

Density of solution = 1.115 g/ mL

Step 3: Calculate mass of the solution

Mass of solution = density ¨volume

Mass of solution = 1.115 g/ mL * 150 mL

Mass of solution = 167.25 grams

Step 4: Calculate mass of sucrose

35 % = 0.35 * 167.25 grams

Mass sucrose = 58.54 grams

Step 5: Calculate mass of water

Mass of water = mass of sample - mass of sucrose

Mass of water = 167.25 - 58.54 = 108.71 grams

Step 6: Calculate volume of water

Volume = mass / density

Volume = 108.71 grams / 1g/ mL

Volume = 108.71 mL = 0.10871 L

The volume of the water is 108.71 mL

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All of the following conditions of STP are true except A. 101.3 kPa B.3.81kPa.L/mol.K C. 24.2 L D. 273.15 K

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The answer is c 24.2l

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3 years ago

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The molar solubility of silver bromide, AgBr in pure water is 0.0007350 mol/L. What is the

gayaneshka [121]

Answer:

0.000000540

Explanation:

Step 1: Make an ICE chart for the solution of AgBr

"S" represents the molar solubility of AgBr

AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)

I 0 0

C +S +S

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Step 2: Write the expression for the solubility product constant (Ksp)

Ksp = [Ag⁺] [Br⁻] = S × S

Ksp = S² = (0.0007350)² = 0.000000540

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