Question

Review Constants 1 Periodic Table An acid solution is 0.120 Min HCl and 0210 Min HSO4 Part A You may want to reference Pages 307-345) Section 8.7 while completing this problem What volume of a 0.160 M solution of KOH must be added to 5500 mL of the acidic solution to completely neutralize all of the add? Express the volume in liters to three significant figures. IVO E RO? V. Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining

Answer 1

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Imot (3) - > n = MXV nuso Flere, the acid provide to lows from till and H₂SO4 . This be ious reacts with oth ious of Kaoll. (ag) + Ort car - H₂o e- Hun hat ear and Ms Of car, ancagst 3040 I mol - Imol 2 mol. 3 Molarity = no. of moles of solute (mol) volume of solution (2) There = Mecox Veicu = (0.120 M) (550.0mL) from equation ③ = (0.120 mol/L) (0.55L) = 0.066 mol Hu: not = lol .. nyt = 0.066 mol SO4 = Mr₂SO4 X V4.S04 = (0. 190 mol/L) (0.55L) = 0.1045 mot! MH SOL at = 1:2 u. Mut= 2x0. 1045 mol = 0.209 mol Now total nut = 0.066 mol + 0.209 mol = 0.275 mol From equation ③ I ml Ht reacts with Imo DH ious . 0.275 mol Ht reacts with 0.275 mol OH and noo: "OH = 1:1 . "Kaoh = 0.275 m of #m Kaol = 0.140 mol/h ; V KOH = ?? MEN = v= m = 0.275 mol → 1.96L 3 Sigulfrcant figures, on equation 0.140 mol = 1.964