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In a titration of a strong base with a strong acid 10mL of a .5M KOH...

In a titration of a strong base with a strong acid 10mL of a .5M KOH were added to a 25mL of a .25M HCl solution. Calculate the pH of the solution after the addition of the KOH.

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Answer #1

KOH and HCl are strong base and strong acid respectivelly. If equal amounts of the two are added they will neutralise and solution will be neutral. However if any one is present in greater quantity he solution will not be neutral.

Moles of KOH added = Molarity X Volume = 0.5 X 10 = 5mmol = 0.005 mol

Because 1mmol = 0.001mol

Moles of HCl = 25 X 0.25 = 6.25 mmol = 0.00625 mol

Hence 0.005 moles of KOH will react with 0.005moles of HCl.

Amount of HCl unreacted = 0.00625 - 0.005 = 0.00125 mole

Total volume = 10mL + 25mL = 35mL = 0.035L

Molarity of Unreacted HHCl = No. of moles / Volume in Liters = 0.00125/0.035 = 0.0357M

0.0357M HCl will ionise to produce H+

pH = - log[H+]

pH = - log[0.0357]

pH = 1.447

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