Question

23. What is the pH at the stoichiometric point for the titration of 50.00 mL of 0.100 M of HNO2(ag) with 0.200 M NaOH(ag)? The value of Ka for HNO2 is 1.0 x 10-7

Answer 1

041 ST9 ANO: Molarity Zolly volume & 50ml = 0.052 pka=7 ka = 1x107 Ipka_ -log ka-log Cixto Man nout molers = 0-1X0-05 = 0.005 plw = 14 The stoichiometric point of atitration is the point at which the number of moles of acid and base are stoichiometrically equal to one another moles of moles of 0.005 Hng Nach moles of wooh Molarity of - volume of wao H (L) Naolt 0.005 0-2 = volume of Naolt volume of NaoH = 0.0252 ANO + Naolt > NaNg tito weakacid strong base salt To.005 0.005 to-005 0.005 -o-oosto Total volume = 0.05+ 0.025 = 0.075L 0.06667M = [] 0.005 ENANG] = 2 = 0.025 For this salt pl: formula. pH = 2 [pfw xpka troge] = £[14+7 +log (0.06667) pl= 9.912

pH formula's for salt's weak acidt strong base ->salft ito pH = ± [pKW + pka + log c] strong acid & Weak base salf tho pH = ₃ [ pkW _pkbe_loge