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A solution prepared by mixing 100.0 ml of 0.400M HCl with 100.0ml of 0.400M NH3 the PH is?
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A solution prepared by mixing 100.0 ml of 0.400M HCl with 100.0ml of 0...
Introduction:
In this problem, we have two solutions: 0.400 M HCl (hydrochloric acid) and 0.400 M NH3 (ammonia). We will mix these solutions together and determine the pH of the resulting solution.
Given:
- Volume of HCl solution = 100.0 ml
- Volume of NH3 solution = 100.0 ml
- Concentration of HCl solution = 0.400 M
- Concentration of NH3 solution = 0.400 M
Step 1: Calculate moles of HCl and NH3:
To find the moles of HCl and NH3, we will use the formula:
Moles = Concentration * Volume (in liters)
Moles of HCl = 0.400 M * (100.0 ml / 1000 ml/L) = 0.0400 moles
Moles of NH3 = 0.400 M * (100.0 ml / 1000 ml/L) = 0.0400 moles
Step 2: Determine the limiting reactant:
Since the moles of HCl and NH3 are equal, neither is in excess, and both will react completely with each other.
Step 3: Write the balanced chemical equation:
HCl + NH3 → NH4Cl
Step 4: Determine the moles of the product:
Since HCl and NH3 react in a 1:1 ratio, the moles of the product (NH4Cl) will also be 0.0400 moles.
Step 5: Calculate the concentration of the resulting solution:
The total volume of the resulting solution is the sum of the volumes of the two solutions: 100.0 ml + 100.0 ml = 200.0 ml = 0.200 L
The concentration of the resulting solution can be calculated as:
Concentration = Moles / Volume
Concentration of the resulting solution = 0.0400 moles / 0.200 L = 0.200 M
Step 6: Calculate the pOH:
To determine the pOH, we need to find the concentration of hydroxide ions (OH-) in the resulting solution. Since NH4Cl is a salt, it will dissociate into NH4+ and Cl- ions. NH4+ is the conjugate acid of NH3 and will react with water to produce NH3 and H3O+ ions.
NH4+ + H2O → NH3 + H3O+
Since HCl is a strong acid, it will dissociate completely into H+ and Cl- ions.
HCl → H+ + Cl-
Therefore, the concentration of H3O+ ions in the resulting solution will be equal to the concentration of HCl, which is 0.200 M.
Since water is neutral, the concentration of OH- ions can be determined using the equation:
[OH-] * [H3O+] = 1.0 x 10^-14
[OH-] = 1.0 x 10^-14 / [H3O+]
[OH-] = 1.0 x 10^-14 / 0.200
Step 7: Calculate
In this problem, we have two solutions: 0.400 M HCl (hydrochloric acid) and 0.400 M NH3 (ammonia). We will mix these solutions together and determine the pH of the resulting solution.
Given:
- Volume of HCl solution = 100.0 ml
- Volume of NH3 solution = 100.0 ml
- Concentration of HCl solution = 0.400 M
- Concentration of NH3 solution = 0.400 M
Step 1: Calculate moles of HCl and NH3:
To find the moles of HCl and NH3, we will use the formula:
Moles = Concentration * Volume (in liters)
Moles of HCl = 0.400 M * (100.0 ml / 1000 ml/L) = 0.0400 moles
Moles of NH3 = 0.400 M * (100.0 ml / 1000 ml/L) = 0.0400 moles
Step 2: Determine the limiting reactant:
Since the moles of HCl and NH3 are equal, neither is in excess, and both will react completely with each other.
Step 3: Write the balanced chemical equation:
HCl + NH3 → NH4Cl
Step 4: Determine the moles of the product:
Since HCl and NH3 react in a 1:1 ratio, the moles of the product (NH4Cl) will also be 0.0400 moles.
Step 5: Calculate the concentration of the resulting solution:
The total volume of the resulting solution is the sum of the volumes of the two solutions: 100.0 ml + 100.0 ml = 200.0 ml = 0.200 L
The concentration of the resulting solution can be calculated as:
Concentration = Moles / Volume
Concentration of the resulting solution = 0.0400 moles / 0.200 L = 0.200 M
Step 6: Calculate the pOH:
To determine the pOH, we need to find the concentration of hydroxide ions (OH-) in the resulting solution. Since NH4Cl is a salt, it will dissociate into NH4+ and Cl- ions. NH4+ is the conjugate acid of NH3 and will react with water to produce NH3 and H3O+ ions.
NH4+ + H2O → NH3 + H3O+
Since HCl is a strong acid, it will dissociate completely into H+ and Cl- ions.
HCl → H+ + Cl-
Therefore, the concentration of H3O+ ions in the resulting solution will be equal to the concentration of HCl, which is 0.200 M.
Since water is neutral, the concentration of OH- ions can be determined using the equation:
[OH-] * [H3O+] = 1.0 x 10^-14
[OH-] = 1.0 x 10^-14 / [H3O+]
[OH-] = 1.0 x 10^-14 / 0.200
Step 7: Calculate
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A solution prepared by mixing 100.0 ml of 0.400M HCl with 100.0ml of 0...
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A solution prepared by mixing 100.0 ml of 0.400M HCl with 100.0ml of 0.400M NH3 the PH is?
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A solution prepared by mixing 100.0 ml of 0.400M HCl with 100.0ml of 0.400M NH3 the PH is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A solution prepared by mixing 100.0 ml of 0.400M HCl with 100.0ml of 0.400M NH3 the PH is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solution prepared by mixing 100.0 ml of 0.400M HCl with 100.0ml of 0.400M NH3 the PH is?.
A solution prepared by mixing 100.0 ml of 0.400M HCl with 100.0ml of 0.400M NH3 the PH is? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A solution prepared by mixing 100.0 ml of 0.400M HCl with 100.0ml of 0.400M NH3 the PH is? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solution prepared by mixing 100.0 ml of 0.400M HCl with 100.0ml of 0.400M NH3 the PH is?.
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