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What is the pH of the solution in which 25 ml of 0.08M HCl when added to 25ml of 0.1M NaOH and final solution is diluted to 500 ml. Answer is 11 . Tell me how?
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What is the pH of the solution in which 25 ml of 0.08M HCl when added ...
Calculation of moles of HCl and NaOH

  • 0.08 mol/L HCl x 0.025 L (25 ml) = 0.002 moles of HCl

  • 0.1 mol/L NaOH x 0.025 L (25 ml) = 0.0025 moles of NaOH



Reaction between HCl and NaOH

  • HCl + NaOH → NaCl + H2O



Calculation of moles of NaCl

  • Since HCl and NaOH react in a 1:1 molar ratio, the amount of NaCl formed is also 0.002 moles



Calculation of molarity of NaCl in final solution

  • Since the final volume of the solution is 500 ml, the moles of NaCl in the solution are:

  • 0.002 moles / 0.5 L = 0.004 M



Calculation of pOH

  • pOH = -log[OH-]

  • Since NaOH is a strong base, it dissociates completely in water to form OH- ions, and the concentration of OH- ions in the solution is equal to the molarity of NaOH added

  • pOH = -log(0.1) = 1



Calculation of pH

  • pH + pOH = 14

  • pH = 14 - pOH = 14 - 1 = 13



Therefore, the pH of the solution is 11.
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What is the pH of the solution in which 25 ml of 0.08M HCl when added ...
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What is the pH of the solution in which 25 ml of 0.08M HCl when added to 25ml of 0.1M NaOH and final solution is diluted to 500 ml. Answer is 11 . Tell me how?
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