50.00 mL of a 0.020 M solution of Ba(OH)₂ is added to 50.00 mL of a 0.060 M solution of HNO₃ - VCE - SSCE Chemistry - Question 6 - 2005 - Paper 1
Question 6
50.00 mL of a 0.020 M solution of Ba(OH)₂ is added to 50.00 mL of a 0.060 M solution of HNO₃.
The hydrogen ion concentration in the resultant solution, in mole per l... show full transcript
Worked Solution & Example Answer:50.00 mL of a 0.020 M solution of Ba(OH)₂ is added to 50.00 mL of a 0.060 M solution of HNO₃ - VCE - SSCE Chemistry - Question 6 - 2005 - Paper 1
Step 1
Calculate moles of Ba(OH)₂
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Answer
First, we calculate the moles of Ba(OH)₂ in the solution:
Molarity (M) = moles/volume (L)
Moles of Ba(OH)₂ = 0.020 M × 0.05000 L = 0.001 Moles.
Step 2
Calculate moles of HNO₃
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Answer
Next, we calculate the moles of HNO₃:
Moles of HNO₃ = 0.060 M × 0.05000 L = 0.003 Moles.
Step 3
Determine NH₄⁺ concentration
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Answer
Ba(OH)₂ dissociates to produce 2 moles of OH⁻ per mole:
Moles of OH⁻ from Ba(OH)₂ = 2 × 0.001 = 0.002 Moles.
Now we analyze the neutralization reaction:
HNO₃ + OH⁻ → H₂O + NO₃⁻.
Thus, moles of HNO₃ reacting with OH⁻ is 0.002 Moles, leaving 0.001 Moles of HNO₃ unreacted.
Step 4
Calculate total volume of the solution
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Answer
The total volume of the solution after mixing:
Total Volume = 50.00 mL + 50.00 mL = 100.00 mL = 0.100 L.
Step 5
Calculate final concentration of H⁺
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Answer
The concentration of unreacted HNO₃ equals the concentration of H⁺ ions:
Molarity (H⁺) = moles / volume (L)
Moles of HNO₃ = 0.001 Moles
Concentration of H⁺ = 0.001 Moles / 0.100 L = 0.010 M.
Step 6
Final answer
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Answer
The hydrogen ion concentration in the resultant solution is 0.010 M. Thus, the answer is A.
