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A 20.0 ml sample of 0.150 m ethylamine is titrated with 0.0981 m hcl. what is the ph after the addition of 5.0 ml of hcl? for ethylamine, pkb = 3.25

Kevin Sutter

in Chemistry

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Kathy Robinson on August 1, 2018

The PKa = 14 - pKbpKa = (14 - 3.25) = 10.75 pH = pKa + log(base/added acid)--> base and the acid in moles.15 M x .02 L = .003 moles Et; .005 M x .0981 = 4.905 x 10^-4 moles of HCl; .003 - 4.905 x 10^-4 = .0025095 rest of moles of EtpH = 10.75 + (log(((.0025095 moles Et) / (4.095 x 10^-4 mol HCl))) =pH = 10.75 + .70895 = 11.46 pH units


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