An analytical chemist is titrating 90.6 mL of a 0.2800 M solution of diethylamine ((C₂H₂)₂NH) with a 0.1100 M solution of HNO3. The pK, of diethylamine is 2.89. Calculate the pH of the base solution after the chemist has added 270.3 mL of the HNO3 solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO3 solution added. Round your answer to 2 decimal places. pH = X S ?

Ag+(aq)+e-→Ag(s) 0.800 Al3+(aq)+3e-→Al(s) -1.676 Au+(aq)+e-→Au(s) 1.692 Au3+(aq)+3e-→Au(s) 1.498 Ba2+(aq)+2e-→Ba(s) -2.912 Be2+(aq)+2e-→Be(s) -1.847 Br2(l)+2e-→2Br-(aq) 1.066 Ca2+(aq)+2e-→Ca(s) -2.868 Cd2+(aq)+2e-→Cd(s) -0.403 Ce4+(aq)+e-→Ce3+(aq) 1.72 Cl2(g)+2e-→2Cl-(aq) 1.358 Co2+(aq)+2e-→Co(s) -0.28 Co3+(aq)+e-→Co2+(aq) 1.92 Cr2+(aq)+2e-→Cr(s) -0.913 Cr3+(aq)+e-→Cr2+(aq) -0.407 Cr3+(aq)+3e-→Cr(s) -0.744 CrO42-(aq)+4H2O(l)+3e-→Cr(OH)3(s)+5OH-(aq) -0.13 Cr2O72-(aq)+14H+(aq)+6e-→2Cr3+(aq)+7H2O(l) 1.36 Cu+(aq)+e-→Cu(s) 0.521 Cu2+(aq)+e-→Cu+(aq) 0.153 Cu2+(aq)+2e-→Cu(s) 0.342 F2(g)+2e-→2F-(aq) 2.866 Fe2+(aq)+2e-→Fe(s) -0.447 Fe3+(aq)+e-→Fe2+(aq) 0.771 Fe3+(aq)+3e-→Fe(s) -0.037 2H+(aq)+2e-→H2(g) 0.000 2H2O(l)+2e-→H2(g)+2OH-(aq) -0.828 H2O2(aq)+2H+(aq)+2e-→2H2O(l) 1.776 Hg2+(aq)+2e-→Hg(l) 0.851 2Hg2+(aq)+2e-→Hg22+(aq) 0.92 Hg22+(aq)+2e-→2Hg(l) 0.797 I2(s)+2e-→2I-(aq) 0.536…

b. C. ΝΗ 19 starting material final product starting material d. N-CH3 ὋΗ final product OH -CH3 $1 starting material final product

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