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aivan3 [116]
1 year ago
13

Calculate the ph of a solution formed by mixing 250. 0 ml of 0. 15 m nh4cl with 200. 0 ml of 0. 12 m nh3. the kb for nh3 is 1. 8

× 10-5.
i. 9. 45
ii. 4. 74
iii. 9. 06
iv. 04. 55
v. 9. 26
Chemistry
1 answer:
Phantasy [73]1 year ago
4 0

The potential of the hydrogen in water gives the pH of the acidity and basicity of the solution. The pH of the solution is 9.06. Thus, option iii is correct.

<h3>What is pH?</h3>

The pH of a substance is given by the subtraction of the pOH from 14 which is the range of the pH scale.

The dissociation reaction for ammonia is given as,

NH₃ (aq) + H₂O (l)  ⇄  NH₄⁺ (aq) + OH⁻ (aq)

Here, the concentration of ammonia is [NH₃] - x, ammonium ion is [NH₄⁺] + x, and hydroxide ion is x.

The molar concentration of ammonia is,

M = (0.12 M × 0.2 L) ÷ 0.45 L = 0.053 M

The molar concentration of ammonium ion is,

M = (0.15 M × 0.25 L) ÷ 0.45 L = 0.083 M

From the base dissociation constant and previous concentration from the reaction, the value of x or hydroxide ion is calculated as,

Kb = [NH₄⁺][OH⁻] ÷ [NH₃]

1.8 × 10⁻⁵ (0.053 - x) - (0.083 + x) × x = 0

x = [OH⁻] = 1.15 x 10⁻⁵

The pH from the hydroxide ion is calculated as,

pOH = - log [OH⁻]

= - log (1.15 x 10⁻⁵)

= 4.94

Further,

pH = 14 - pOH

= 14 - 4.94

= 9.06

Therefore, option iii. 9.06 is the pH of the solution.

Learn more about pH here:

brainly.com/question/14790520

#SPJ4

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siniylev [52]

Answer:

5.4 M.

Explanation:

  • At complete neutralization: It is known that the no. of millimoles of acid equal that of the base.

<em>(MV)acid = (MV)NaOH</em>

M of acid = ??? M, V of acid = 35.0 mL.

M of NaOH = 3.0 M, V of NaOH = 63.0 mL.

∴ M of acid = (MV)NaOH / (V)acid = (3.0 M)(63.0 mL)/(35.0 mL) = 5.4 M.

4 0
3 years ago
A solution is prepared by dissolving ammonium sulfate in enough water to make of stock solution. A sample of this stock solution
Pani-rosa [81]

Answer: molarity of ammonium ions = 0.274mol/L

molarity of sulfate ions = 0.137mol/L

<em>Note: The complete question is given below</em>

A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00-mL sample of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

Explanation:

Molar concentration = no of moles/volume in liters

no of moles = mass/molar mass

mass of ammonium sulfate = 10.8g, molar mass of ammonium sulfate, (NH₄)₂SO₄ = (14+4)*2 + 32+ (16)*4 = 132g/mol

no of moles = 10.8g/132g/mol = 0.0820moles

<em>Molarity of stock solution = 0.0820mol/(100ml/1000ml* 1L) = 0.0820mol/0.1L Molarity of stock solution = 0.820mol/L</em>

Concentration of final solution is obtained from the dilution formula,

<em>C1V1 = C2V2</em>

C1 = 0.820M, V1 = 10mL, C2 = ?, V2 = 60mL

C2 = C1V1/V2

C2 = 0.820*10/60 = 0.137mol/L

molar concentration of ions = molarity of solution * no of ions

molarity of ammonium ions = 0.137mol/L * 2 = 0.274mol/L

molarity of sulfate ions = 0.137 mol/L * 1 = 0.137mol/L

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3 years ago
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3 years ago
What is the molarity of a solution in which 0.321 g of calcium chloride is dissolved in 1.45 L water?
lana66690 [7]

Answer:

Molarity = 0.002 M

Explanation:

Given data:

Mass of calcium chloride = 0.321 g

Volume of water = 1.45 L

Molarity of solution = ?

Solution:

Molarity = number of moles / volume in litter.

We will calculate the number of moles of calcium chloride first.

Number of moles = mass/molar mass

Number of moles = 0.321 g/ 110.98 g/mol

Number of moles = 0.003 mol

Molarity:

Molarity = 0.003 mol / 1.45 L

Molarity = 0.002 M

7 0
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The average atomic mass recorded on the periodic table for cobalt is 58.933 u. This indicates that the most abundant isotope of
OleMash [197]
The average atomic mass of the element is the sum of the products of the percentage abundance of isotope and its mass number. Therefore, for atomic mass equal to 58.933, the most abundant isotope is cobalt-59. Thus, the answer is letter C. 
7 0
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