Question

50.0 mL of 0.150 M HNO₂ is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

Answer 1

Titration of Weak acid with strong base molarity of HNO2 = 0.150 M volume of HNO2 = 50.00 mL = 0.0500 L molarity of NaOH = 1.000 M Ka of HNO2 = 4.0x104 pka = -log[Ka] = -log[4.0*104 ] = 3.40 Macid*Vacid = Mbase*Vbase 0.150 0.0500 = 1.000 * Vbase volume of base NaOH = 0.0075 L total volume = 0.0500 + 0.0075 = 0.0575 L Calculating initial number of moles: moles = Molarity * Volume number of moles of HNO2 = 0.150 x 0.05000 = 0.0075 mol At equivalnece point only salt remains moles of acid = base = salt = 0.0075 mol HNO2 + NaOH → NaNO2 + H2O moles of salt NaNO2 = 0.0075 mol Calculating Concentration : concentration = moles / Volume concentration of NaNO2 = 0.0075 / 0.0575 = 0.1304 M Calculating pH: pH = [14+ pka +logC ]/2 pH = [14 + 3.40 +log ( 0.1304 )]/2 pH = 8.26