Question

You may want to reference (Pages 795 - 812) Section 17.4 while completing this problem. A 42.0 mL sample of 0.120 M HNO, is titrated with 0.256 M KOH (K, for HNO, is 4.57x10-4) Part A Determine the pH at the equivalence point for the titration of HNO, and KOH. VAE ? pH = Submit Request Answer Provide Feedback 29

Answer 1

Answer Neutralization reaction: HNO2 + KOH KNO2 + H₂O At the equivalence point, HNO2 will be completely converted into NOZ. Number of mol of HNO2 in 12.0 mL (0.0420 L) of 0.120M HNO2 (0.120x 0.0420) mot 5.04x 103 mol Hence, at the equivalence point, 5.04x 103 mot of HNO₂ will be converted to 5.04x 103 mol of Noz. Volume of Koh needed to reach equivalence point (volume of HNO2 (motarity of HN02) (molarity of KOH) (12-0 mL) x (0-120 M) (0.256M) 19.7 mL 0.0197 L Total volume of mixture at equivalence point (0.0920 +0.0197) L 0.0617 L Hence, concentration of Noz at the equivalence point. 5.04x103 = 0.0817 M construct an ICE table corresponding to hydrolysis of Noz to calculate pH at the equivalence point. Noz + H₂O Y HNO₂+ oH I(M): 0.0817 C(M): -X E(M): 0.0817-8 [HNO] [OH-] [Noz 4.57X101 x² 2.19x 10" 0.0817-X Ny x +(2-19x17")x - (1-79x1012) - - (2-19x101)+. +1(2.19X10'')? +(4x1x1.79x102) (2x1) = 1.34 x 100 M M 0.0617 0 tx +2 Х 2014 kb (NO2) = Ka (HNO2) 1019

(ax'+bx+c = 0 >> X- - b ± √b²-qac ; here axo) za PH = 14-POH • 14+ log[OH- = 14+ logx = 14+ log (1-34 * 10%) = 8.13 So, PH at the equivalence point is 18.13