Question

What volume, in milliliters, of 0.150 M NaOH should be added to a 0.130 L solution of 0.028 M glycine hydrochloride (pKal = 2.350, pKa2 = 9.778) to adjust the pH to 2.957 NaOH volume = 10.39

Answer 1

H-c-coon - TNH3 Glycine At pH = 2.95 glycine exist as follow loka, =2.35 H-C-COOH H-E-wo- IN43 TNH₂ (acidic form) Conjugare base So n'ячne Solution aq a bag eh solution - vko uul m'ni очдог and mol of Maoh added = moly conjugate bare formed let malgraoh added = x mol of Glycine present = nolarity X volume = 0.028 mol X 0.130% = 3.64 mol Nach

NaoH + Willoll n-c-cool HC-coon ENKS initial initial i 3.64 change Final 3.64 x

using Henderson Hasselbach equation PH=pkatlog [base] Cacid] 2.95= 2.35 t log (3) 3.64-2 0.6 = log (x) "364-2 = 2 3.98 C 3.64-2 14.5 - 3.98 x = x 14.5 = 4.982 2.91 = se mol of NaOH = d.91 mol added Molarity of Naol = 0.150m volume of Nach = mol molarity = 2.91 mol = 19.um Olsom612-1 volume I haoh added = 19.4 mL Answee