Question

Ka of (NH4)+ : 5.70x10^-10

Ka of Piperidinium ion: 7.50x10^-12

Ka of ethyl ammonium ion: 3.18x10^-10

Ka of anilinium ion: 2.51x10^-5

What is the pH of a solution that is (This problem requires values in your textbook's specific appendices, which you can access through the OWLv2 MindTap Reader. You should not use the OWLv2 References' Tables to answer this question as the values will not match.) a prepared by dissolving 4.30 g of (NH4),S04 in water, adding 100.0 mL of 0.2906 M NaOH, and diluting to 800.0 mL? pH = b 0.146 M in piperidine and 0.070 M in its chloride salt? pH = C 0.083 M in ethylamine and 0.136 M in its chloride salt? pH = d prepared by dissolving 3.05 g of aniline (93.13 g/mol) in 100 mL of 0.0630 M HCl and diluting to 850.0 mL? pH =

Answer 1

132.14 kn of NH2 = & Ko of 10th 1,75 x 10-5 NH3 = 5,70x10 10 Por log (1175x105) = 4.757. mole of NHG = 4.30 x 2 = 0.0651 mole mole of NaOH = 0.100 L x 0.2906 moto/= 0.02906 mole. POH= pky & log 1 Salt? (base] * = 4.752 & log (6.0651- 0.02906) 0.02906 poH = 4.8 pH = 14 - 4.85 TpH = 9.157 6 4 = -log ky = -log (10-14 12) = 2.875 pOH = 2.875 + log (0.070) pott a 2.556 pH = 14-2.556 (pH = 11:44 © pres - - Tento) = 4050 pOH = 4.50 10.083 pott = 4.71 PH a 14-4.21 pub=lez (10-1415) = 9.90 pH = 9:29) mole of aniline pot = 9.40 (0.0063 ne 93-13 logy (0-0524-010062)5 pot = 8.78 0.0327-0.0063 = 0.0327 pH = 14-8.7 Mole of Her Owx 0.0630 (0.136 STX 105 mole 4x2 -0.0063 mote