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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  1. cn u help

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  2. 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

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