Question

Find the pH of the equivalence point and the volume (mL) of 0.0558 M KOH needed to reach the equivalence point in the titration of 23.4 mL of 0.0390 M HNO2.

Answer 1

pH formula's for salt's weak acidt strong base - salft ito pH = ₃ [pkw + pka + loge] strongacid & weakbase pt = ₃ [ pk w _pk_loge salf t o

Kaob HNO = 4. 1 pka-log ka HNO: Molarity = 0.039M volume=23-4m1=0.02342 M= 4 no ot moles = 0.039 x 0.0234 = 0.0009126 at equivalence point moles of moles of pla, -log (4.5x10t) = 3-347 I pkw=14 0.000 9126 ANG - KOH moles of kolt molarity of Kot volume of to 0.0009126 6.65582 volume of Kolt 16.355m) 0.0 163552= volume of kolt= nobate salt T Hwo + kot - kNot to weak acid strong base I 0.0009126 0.0009126 to c/-0.0004126 1-8-0009126 +0.0009126 0.000 9126 Total volume = o.0234+ 0.016355- 0-0009126 0.022956 m = [c7 234+0.016355=0.0397552 - 0.022951 I KNQ] = 2 = 0.039755 For this salt.ph Formula pH = + [pkw toka tloge] pH = $ {14 + 3.347 +log (0.022956) pH 7.854