Question

In a titration of 200 ml of 0.150 M HCl with 0.200 M KOH, what is the pH after 70.0 ml of KOH have been added?

Answer 1

The neutralization reaction between HCl and KOH is,

HCl (aq) + KOH (aq) KCl (aq) + H₂O

1 mole of HCl 1 mole of KOH. ......... (1)

For HCl,

Molarity = 0.150 M, Volume = 200.0 mL = 0.200 L

# of moles of H+ = # of moles of HCl = Molarity * Volume in L = 0.150 * 0.200 = 0.030 mol.

For KOH,

Molarity = 0.20 M, Volume = 70.0 mL = 0.07 L

# of moles of HO- = # of moles of KOH = 0.20 * 0.07 = 0.014 mole

By mole relationship its clear that,

# of moles of HCl left unreacted = 0.030 - 0.014 = 0.016 mol.

Total volume after mixing = 200.0 + 70.0 = 270.0 mL = 0.270 L

Then,

[H⁺] = # of moles of H⁺ / Volume of solution in L = 0.016 mol / 0.270 L = 0.059 M

pH = -log[H⁺] = -log (0.059) = 1.23

pH of the solution is 1.23

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