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pOH of a solution formed by mixing 60 ml of 0.2 M NaCN (aq) with 80 ml of 0.1 M HCl (aq) will be
(Given : Ka(HCN) = 5 × 10–10; log 2 = 0.3; log 5 = 0.7)
(Given : Ka(HCN) = 5 × 10–10; log 2 = 0.3; log 5 = 0.7)
- a)9
- b)7
- c)5
- d)3
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
pOH of a solution formed by mixing 60 ml of 0.2 M NaCN (aq) with 80 ml...
= 10 – log 5 – log 2
pH = 9; pOH = 5
Most Upvoted Answer
pOH of a solution formed by mixing 60 ml of 0.2 M NaCN (aq) with 80 ml...
Understanding the Reaction
When mixing NaCN and HCl, a reaction occurs between the weak base (CN-) and the strong acid (HCl). The primary reaction is:
- HCl + CN- → HCN + Cl-
Calculating Moles of Reactants
- Moles of NaCN:
Volume = 60 ml = 0.060 L
Concentration = 0.2 M
Moles of NaCN = 0.060 L × 0.2 mol/L = 0.012 mol
- Moles of HCl:
Volume = 80 ml = 0.080 L
Concentration = 0.1 M
Moles of HCl = 0.080 L × 0.1 mol/L = 0.008 mol
Determining Limiting Reactant
Since HCl is in lesser moles (0.008 mol) compared to NaCN (0.012 mol), HCl is the limiting reactant.
Calculating Remaining Moles
After the reaction:
- Moles of NaCN remaining = 0.012 mol - 0.008 mol = 0.004 mol
- Moles of HCN formed = 0.008 mol
Final Concentration in the Mixture
Total volume after mixing = 60 ml + 80 ml = 140 ml = 0.140 L
- Concentration of CN-:
[CN-] = 0.004 mol / 0.140 L = 0.02857 M
- Concentration of HCN:
[HCN] = 0.008 mol / 0.140 L = 0.05714 M
Calculating pKa and pH
Using the relationship of weak acid and its conjugate base:
- Ka (HCN) = 5 × 10^–10
- pKa = -log(Ka) = -log(5) + 10 = 7 - 0.7 = 9.3
Using the Henderson-Hasselbalch equation:
- pH = pKa + log([A-]/[HA])
- pH = 9.3 + log(0.02857/0.05714)
- pH = 9.3 + log(0.5) = 9.3 - 0.3 = 9.0
Calculating pOH
- pOH = 14 - pH = 14 - 9 = 5
Thus, the pOH of the solution is 5, confirming that the correct answer is option 'C'.
When mixing NaCN and HCl, a reaction occurs between the weak base (CN-) and the strong acid (HCl). The primary reaction is:
- HCl + CN- → HCN + Cl-
Calculating Moles of Reactants
- Moles of NaCN:
Volume = 60 ml = 0.060 L
Concentration = 0.2 M
Moles of NaCN = 0.060 L × 0.2 mol/L = 0.012 mol
- Moles of HCl:
Volume = 80 ml = 0.080 L
Concentration = 0.1 M
Moles of HCl = 0.080 L × 0.1 mol/L = 0.008 mol
Determining Limiting Reactant
Since HCl is in lesser moles (0.008 mol) compared to NaCN (0.012 mol), HCl is the limiting reactant.
Calculating Remaining Moles
After the reaction:
- Moles of NaCN remaining = 0.012 mol - 0.008 mol = 0.004 mol
- Moles of HCN formed = 0.008 mol
Final Concentration in the Mixture
Total volume after mixing = 60 ml + 80 ml = 140 ml = 0.140 L
- Concentration of CN-:
[CN-] = 0.004 mol / 0.140 L = 0.02857 M
- Concentration of HCN:
[HCN] = 0.008 mol / 0.140 L = 0.05714 M
Calculating pKa and pH
Using the relationship of weak acid and its conjugate base:
- Ka (HCN) = 5 × 10^–10
- pKa = -log(Ka) = -log(5) + 10 = 7 - 0.7 = 9.3
Using the Henderson-Hasselbalch equation:
- pH = pKa + log([A-]/[HA])
- pH = 9.3 + log(0.02857/0.05714)
- pH = 9.3 + log(0.5) = 9.3 - 0.3 = 9.0
Calculating pOH
- pOH = 14 - pH = 14 - 9 = 5
Thus, the pOH of the solution is 5, confirming that the correct answer is option 'C'.
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pOH of a solution formed by mixing 60 ml of 0.2 M NaCN (aq) with 80 ml of 0.1 M HCl (aq) will be(Given : Ka(HCN) = 5 × 10–10; log 2 = 0.3; log 5 = 0.7)a)9b)7c)5d)3Correct answer is option 'C'. Can you explain this answer?
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pOH of a solution formed by mixing 60 ml of 0.2 M NaCN (aq) with 80 ml of 0.1 M HCl (aq) will be(Given : Ka(HCN) = 5 × 10–10; log 2 = 0.3; log 5 = 0.7)a)9b)7c)5d)3Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about pOH of a solution formed by mixing 60 ml of 0.2 M NaCN (aq) with 80 ml of 0.1 M HCl (aq) will be(Given : Ka(HCN) = 5 × 10–10; log 2 = 0.3; log 5 = 0.7)a)9b)7c)5d)3Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for pOH of a solution formed by mixing 60 ml of 0.2 M NaCN (aq) with 80 ml of 0.1 M HCl (aq) will be(Given : Ka(HCN) = 5 × 10–10; log 2 = 0.3; log 5 = 0.7)a)9b)7c)5d)3Correct answer is option 'C'. Can you explain this answer?.
pOH of a solution formed by mixing 60 ml of 0.2 M NaCN (aq) with 80 ml of 0.1 M HCl (aq) will be(Given : Ka(HCN) = 5 × 10–10; log 2 = 0.3; log 5 = 0.7)a)9b)7c)5d)3Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about pOH of a solution formed by mixing 60 ml of 0.2 M NaCN (aq) with 80 ml of 0.1 M HCl (aq) will be(Given : Ka(HCN) = 5 × 10–10; log 2 = 0.3; log 5 = 0.7)a)9b)7c)5d)3Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for pOH of a solution formed by mixing 60 ml of 0.2 M NaCN (aq) with 80 ml of 0.1 M HCl (aq) will be(Given : Ka(HCN) = 5 × 10–10; log 2 = 0.3; log 5 = 0.7)a)9b)7c)5d)3Correct answer is option 'C'. Can you explain this answer?.
Solutions for pOH of a solution formed by mixing 60 ml of 0.2 M NaCN (aq) with 80 ml of 0.1 M HCl (aq) will be(Given : Ka(HCN) = 5 × 10–10; log 2 = 0.3; log 5 = 0.7)a)9b)7c)5d)3Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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