Question

A 1 liter solution contains 0.288 M nitrous acid and 0.384 M potassium nitrite. Addition of 0.317 moles of sodium hydroxide will: (Assume that the volume does not change upon the addition of sodium hydroxide.) Raise the pH slightly Lower the pH slightly Raise the pH by several units Lower the pH by several units Not change the pH Exceed the buffer capacity

Answer 1

0.384 CANDIDATE ANSWER 4NO2 I KNO2 ph=pkat logo 07255 0.28879 0.384 S - 3.15 t 109133 la = 3.15 - 3.15+ 0.12 Naoh (01317 Mole:) = 3. 27 before ph HNO + Naon Na Nogt no 0,288 0,317, ·317_05288 be cause insolm Strong Base in present. So [or] come from 0.3) 70.288 Strong Basse- for) = 0.025 pon= -loglou I so raise the ph by / рои 2 - - о: 029 l Several unit pon = 1.53 pu= 14-1:53 n = 12 :46 o hookato 10,468 pH = pke + boy 0.45 ph 27.50+ log 0.75 pH <7,53 0.12 pn = 7:40 Befor pH pk_7153

kolou к со - коч као - Но, 0-416 0.457 0 0 .452-1416 0.416 because it is come from Strong acid, 0.457 = 0.041 so [ht] = 0.041 ph = -log(ne) pc-logo.041 put = 1.38 → After pt So lower the ph by Sevaral units. ③ qka=4.74 c 4.74 enzoon/ CH₂ cook 0.479m 0.359. ph=pkatlong as pn = 4.74+ log 0.74 pn = 4.74 -0.13 pn a 4.60 Before

HIL0180) си,соор, ит о: 3ss o-) &o си, сооп + KI 1 б.) go to: 129 Acidia Buftes о: 3 Ss — O• 180 е" - р-н даа) — + 4 + 7-2-3-4% эч + ': 135 — чэч + Дзе - 21 Lower the ph Slightly , ч. 2y - o. S6 - ч. ) » р ри 2
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