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2 years ago

Calculate the mass of ammonium sulfide (nh4)2s in 3.00 l of a 0.0200 m solution. g (nh4)2s

1 answer:

5 0

Taking into account the definition of molarity, 4.08 grams of ammonium sulfide are present in 3.00 L of a 0.0200 M solution.

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

$Molarity=\frac{number of moles}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$ .

<h3>Mass of ammonium sulfide</h3>

In this case, you know:

Molarity: 0.0200 M
number of moles: ?
volume: 3 L

Replacing in the definition of molarity:

$0.0200 \frac{moles}{liter}=\frac{number of moles}{3 L}$

Solving:

0.0200 $\frac{moles}{liter}$ × 3 L= number of moles

0.06 moles= number of moles

On the other side, the molar mass is the amount of mass that a substance contains in one mole. In this case, the molar mass of ammonium sulfide is 68 g/mole.

Then, the amount of mass contained in 0.06 moles is calculated as:

0.06 moles× $\frac{68 grams}{1 mole}$ = 4.08 grams

Finally, 4.08 grams of ammonium sulfide are present in 3.00 L of a 0.0200 M solution.

Learn more about

molarity:

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molar mass:

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Answer:

(a) pH = 12.73

(b) pH = 10.52

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

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C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

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pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

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(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

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