PopQuiz-3-8 - [email protected]” 3/g/07.9 1 A 25.0 mL sample of an...

Unformatted text preview: 70;) @102.” 3/g/07 .9 1. A 25.0 mL- sample of an acetic acid solution is titrated with a 0.175 M NaOH ‘ solution. The equivalence point is reached when 37.5 mL of the base is added. What is the Eoncentration of acetic acid in the original sample? ' ' @337; Lav/ewflflfiw WW) I ”5%: [0.054 #bfleJK/KM 19%) I . 2 Mei/mflmwl’”). zigzflfifiié (0.;2z5)‘ ~ ' 2. Calculate the pH in the soluy'bg formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzo'c acid (HC7H502, K, = 6.3 X 106)." 7526433 / )4 __ ' g .0 ”,1 male, gammy : :9. m m we weak as? WM WWLMWOZW/agz/mo mm W name #1 57...... ., 34:7 ' 0,5?)nmfléI: 620/0,” $420 fww b $31.0 ML . .. ' 3. Calculate the pH in the solution formed 1 .100 M HCl to , 20.0 mL of 0.100 MNH3. Kb for NH3 is 1.8 x105. p [Q 3 6;] 76/ {momL fiétf)(fl,/0a Mm): MD mm/g ficgg (Maw/W902i” M ”3) ‘9'” ”WW5 . , /. M a/ ' . . [Mat/2+ Dix/Vii; 3:9,": ‘2: 0.023317 ' 0M3} '7’ £30: 51:0 :2 0,3353/‘4 DH; ( .. YflWEzg 0,103 3 P fib+¢éy£fl$§£*%7¢* fl): M %7% ...
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