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The pH of a solution prepared by mixing 2M, 100ml HCl &1M, 200ml NaOH at 25C is Option: (A) 8 (B) 7 (C) 4 (D) 5 The correct option is(B). Explain it.?
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The pH of a solution prepared by mixing 2M, 100ml HCl &1M, 200ml NaOH ...
Since 100 mL of 2 M HCl is 0.2 Moles of the HCl and 200 mL of 1 M NaOH is 0.2 Moles of the NaOH. Also, HCl is a strong acid and NaOH is a strong base, then theoretically this mixture will produce 0.2 Moles of NaCl as the HCl would react with the NaOH to produce NaCl. so that the required solution will be neutral that is pH=7.
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The pH of a solution prepared by mixing 2M, 100ml HCl &1M, 200ml NaOH ...
Calculating the pH of the solution:
To determine the pH of the solution, we first need to calculate the moles of HCl and NaOH present in the solution.

Step 1: Calculate moles of HCl
Moles of HCl = Molarity x Volume (in liters)
Moles of HCl = 2 mol/L x 0.1 L
Moles of HCl = 0.2 moles

Step 2: Calculate moles of NaOH
Moles of NaOH = Molarity x Volume (in liters)
Moles of NaOH = 1 mol/L x 0.2 L
Moles of NaOH = 0.2 moles

Step 3: Determine the limiting reagent
Since both HCl and NaOH have the same number of moles, NaOH is the limiting reagent as it is in lesser amount.

Step 4: Calculate the excess NaOH
Excess NaOH = Moles of NaOH - Moles of HCl
Excess NaOH = 0.2 moles - 0.2 moles
Excess NaOH = 0 moles

Step 5: Calculate the concentration of OH⁻ ions
Concentration of OH⁻ ions = Moles of NaOH / Total volume of solution
Concentration of OH⁻ ions = 0.2 moles / (0.1 L + 0.2 L)
Concentration of OH⁻ ions = 0.2 moles / 0.3 L
Concentration of OH⁻ ions = 0.67 M

Step 6: Calculate pOH
pOH = -log[OH⁻]
pOH = -log(0.67)
pOH ≈ 0.18

Step 7: Calculate pH
pH = 14 - pOH
pH = 14 - 0.18
pH ≈ 13.82
Therefore, the pH of the solution prepared by mixing 2M HCl and 1M NaOH is approximately 13.82, which is close to option (B) 7.
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The pH of a solution prepared by mixing 2M, 100ml HCl &1M, 200ml NaOH at 25C is Option: (A) 8 (B) 7 (C) 4 (D) 5 The correct option is(B). Explain it.?
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