Calculate the pH of a mixture of 20.0 mL of 0.150 M HCl and 10.0 mL of 0.300 M NaOH.[7.00]
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20ml(0.150M HCl) + 10ml(0.0300M NaOH
=> 0.003mole HCl + 0.003mole NaOH
=> (0.003mol/0.03L)HCl + (0.003mol/0.01L)NaOH
=> 0.30M HCl + 0.30M NaOH
=> 0.30M NaCl + 0.30M HOH
Since neither Na^+ or Cl^- will undergo hydrolysis (rxn with water), the outcome of the reaction is dependent upon the autoionization of water only. That is,
HOH <=> H^+ + OH^-
=> [H^+] = [OH^-] = 1 x 10^-7
=> pH = -log[H^+] = -log(1 x 10^-7) = 7.00
=> 0.003mole HCl + 0.003mole NaOH
=> (0.003mol/0.03L)HCl + (0.003mol/0.01L)NaOH
=> 0.30M HCl + 0.30M NaOH
=> 0.30M NaCl + 0.30M HOH
Since neither Na^+ or Cl^- will undergo hydrolysis (rxn with water), the outcome of the reaction is dependent upon the autoionization of water only. That is,
HOH <=> H^+ + OH^-
=> [H^+] = [OH^-] = 1 x 10^-7
=> pH = -log[H^+] = -log(1 x 10^-7) = 7.00