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moles of HCl = 50*0.1 = 5 moles of NaOH = 20.5*0.2 = 4.1 so moles of H+ left
after titration = 5-4.1 = 0.9 so [H+] = 0.9/20.5+50 = 0.0127 so pH = -log([H+]) = - log(0.0127)
=1.896 ANSWER
Solution
moles of HCl = 50*0.1 = 5 moles of NaOH = 20.5*0.2 = 4.1 so moles of H+ left
after titration = 5-4.1 = 0.9 so [H+] = 0.9/20.5+50 = 0.0127 so pH = -log([H+]) = - log(0.0127)
=1.896 ANSWER

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moles of HCl = 500.1 = 5 moles of NaOH = 20.50..pdf

  • 1. moles of HCl = 50*0.1 = 5 moles of NaOH = 20.5*0.2 = 4.1 so moles of H+ left after titration = 5-4.1 = 0.9 so [H+] = 0.9/20.5+50 = 0.0127 so pH = -log([H+]) = - log(0.0127) =1.896 ANSWER Solution moles of HCl = 50*0.1 = 5 moles of NaOH = 20.5*0.2 = 4.1 so moles of H+ left after titration = 5-4.1 = 0.9 so [H+] = 0.9/20.5+50 = 0.0127 so pH = -log([H+]) = - log(0.0127) =1.896 ANSWER