Question
Calculate the pH of a solution obtained by mixing 100ml of 0.2M HCl with 100ml of 0.1M NaOH
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Answer to a math question Calculate the pH of a solution obtained by mixing 100ml of 0.2M HCl with 100ml of 0.1M NaOH
Adonis
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Step 1: Calculate the moles of HCl and NaOH used in the solution.
Moles of HCl = concentration (M) Γ volume (L) = 0.2 mol/L Γ 0.1 L = 0.02 mol
Moles of NaOH = concentration (M) Γ volume (L) = 0.1 mol/L Γ 0.1 L = 0.01 mol
Step 2: Determine which reactant is in excess.
Since the moles of HCl (0.02 mol) are greater than the moles of NaOH (0.01 mol), HCl is in excess.
Step 3: Calculate the moles of HCl remaining after the reaction.
Moles of HCl remaining = moles of HCl initially - moles of NaOH used
Moles of HCl remaining = 0.02 mol - 0.01 mol = 0.01 mol
Step 4: Calculate the concentration of HCl in the solution after the reaction.
The volume of the solution is 100 mL (0.1 L) + 100 mL (0.1 L) = 0.2 L
Concentration of HCl = moles of HCl remaining / volume of solution
Concentration of HCl = 0.01 mol / 0.2 L = 0.05 M
Step 5: Calculate the pOH of the solution.
pOH = -log[OH-]
Since the moles of NaOH added were 0.01 mol and the volume of the solution is 0.2 L, the concentration of OH- is:
Concentration of OH- = moles of OH- / volume of solution = 0.01 mol / 0.2 L = 0.05 M
pOH = -log(0.05) β 1.30
Step 6: Calculate the pH of the solution.
pH + pOH = 14
pH = 14 - pOH = 14 - 1.30 β 12.70
Answer: The pH of the solution obtained by mixing 100 mL of 0.2 M HCl with 100 mL of 0.1 M NaOH is approximately 12.70.
Moles of HCl = concentration (M) Γ volume (L) = 0.2 mol/L Γ 0.1 L = 0.02 mol
Moles of NaOH = concentration (M) Γ volume (L) = 0.1 mol/L Γ 0.1 L = 0.01 mol
Step 2: Determine which reactant is in excess.
Since the moles of HCl (0.02 mol) are greater than the moles of NaOH (0.01 mol), HCl is in excess.
Step 3: Calculate the moles of HCl remaining after the reaction.
Moles of HCl remaining = moles of HCl initially - moles of NaOH used
Moles of HCl remaining = 0.02 mol - 0.01 mol = 0.01 mol
Step 4: Calculate the concentration of HCl in the solution after the reaction.
The volume of the solution is 100 mL (0.1 L) + 100 mL (0.1 L) = 0.2 L
Concentration of HCl = moles of HCl remaining / volume of solution
Concentration of HCl = 0.01 mol / 0.2 L = 0.05 M
Step 5: Calculate the pOH of the solution.
pOH = -log[OH-]
Since the moles of NaOH added were 0.01 mol and the volume of the solution is 0.2 L, the concentration of OH- is:
Concentration of OH- = moles of OH- / volume of solution = 0.01 mol / 0.2 L = 0.05 M
pOH = -log(0.05) β 1.30
Step 6: Calculate the pH of the solution.
pH + pOH = 14
pH = 14 - pOH = 14 - 1.30 β 12.70
Answer: The pH of the solution obtained by mixing 100 mL of 0.2 M HCl with 100 mL of 0.1 M NaOH is approximately 12.70.
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