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picture and insert as an image into this document. If the TA deems your handwriting illegible it is their right to not grade it. When you are finished please save this word document as a PDF and upload to LabArchives Honor Pledge: I pledge, on my honor, that I have neither given, nor received, any unauthorized assistance on this assignment. Signed: Date: BUFFERS 1. (8 points) Fill in the data tables below with your experimental results. For part a, provide the calculation for the 1:9 concentrations of [conj. base] and [conj. acid]. Buffer system: #4 1:1 rto 0.500M acetic acid and 0.500M sodium acetate Preparation of buffer.solutions, Report concentrations in molarity. VolumeRatio-T1conj.base](NDT Iconj.acid|(M)Tlog( con, baselfe nj.acid])AveragepH1 conj.base: conj acid 2:8 8:2 5.75 5.33 4.585 4.08 3.78 91 Calculation for 1:9:Explanation / Answer
molarity = moles / solution in liter
acetic acid = 0.5 M = 0.5 moles in 1 liter solution and sodium acetate = 0.5 M = 0.5 moles in 1 liter solution
1:9 ratio
number of moles in 1 ml = 10-3 L acetic acid = (0.5 moles /L) * 10-3 L = 0.5 * 10-3 moles
total volume of the solution = 1+9 = 10 ml = 10*10-3 L
therefore, [conj. base] = 0.5 * 10-3 moles / 10*10-3 L = 0.05 M
similarly molarity of 9 ml = 9*10-3 L sodium acetate = (0.5 moles/L) * 9*10-3 L = 9* 0.5 * 10-3 moles
total volume of the solution = 1+9 = 10 ml = 10*10-3 L
therefore, [conj. acid] = 9* 0.5 * 10-3 moles / 10*10-3 L = 0.45 M
hence, log [conj. base] / [conj. acid] = log(0.05/0.45) = - 0.0954
similarly,
for, 2:8
number of moles in 2 ml = 2*10-3 L acetic acid = (0.5 moles /L) * 2*10-3 L = 0.5 * 2*10-3 moles
total volume of the solution = 1+9 = 10 ml = 10*10-3 L
therefore, [conj. base] = 0.5 * 2*10-3 moles / 10*10-3 L = 0.1 M
similarly molarity of 8 ml = 9*10-3 L sodium acetate = (0.5 moles/L) * 8*10-3 L = 8* 0.5 * 10-3 moles
total volume of the solution = 1+9 = 10 ml = 10*10-3 L
therefore, [conj. acid] = 8* 0.5 * 10-3 moles / 10*10-3 L = 0.40 M
hence, log [conj. base] / [conj. acid] = log(0.10/0.40) = - 0.0602
for, 1:1
number of moles in 1 ml = 1*10-3 L acetic acid = (0.5 moles /L) * 1*10-3 L = 0.5 * 1*10-3 moles
total volume of the solution = 1+9 = 10 ml = 10*10-3 L
therefore, [conj. base] = 0.5 * 1*10-3 moles / 10*10-3 L = 0.05 M
similarly molarity of 1 ml = 1*10-3 L sodium acetate = (0.5 moles/L) * 1*10-3 L = 1* 0.5 * 10-3 moles
total volume of the solution = 1+9 = 10 ml = 10*10-3 L
therefore, [conj. acid] = 1* 0.5 * 10-3 moles / 10*10-3 L = 0.05 M
hence, log [conj. base] / [conj. acid] = log(0.05/0.05) = 0
for, 8:2
number of moles in 8 ml = 8*10-3 L acetic acid = (0.5 moles /L) * 8*10-3 L = 0.5 * 8*10-3 moles
total volume of the solution = 1+9 = 10 ml = 10*10-3 L
therefore, [conj. base] = 0.5 * 8*10-3 moles / 10*10-3 L = 0.40M
similarly moles of 2 ml = 2*10-3 L sodium acetate = (0.5 moles/L) * 2*10-3 L = 2* 0.5 * 10-3 moles
total volume of the solution = 1+9 = 10 ml = 10*10-3 L
therefore, [conj. acid] = 2* 0.5 * 10-3 moles / 10*10-3 L = 0.1 M
hence, log [conj. base] / [conj. acid] = log(0.40/0.10) = 0.0602
for, 9:1
number of moles in 9 ml = 9*10-3 L acetic acid = (0.5 moles /L) * 9*10-3 L = 0.5 * 9*10-3 moles
total volume of the solution = 1+9 = 10 ml = 10*10-3 L
therefore, [conj. base] = 0.5 * 9*10-3 moles / 10*10-3 L = 0.45M
similarly moles of 1 ml = 1*10-3 L sodium acetate = (0.5 moles/L) * 1*10-3 L = 1* 0.5 * 10-3 moles
total volume of the solution = 1+9 = 10 ml = 10*10-3 L
therefore, [conj. acid] = 1* 0.5 * 10-3 moles / 10*10-3 L = 0.05 M
hence, log [conj. base] / [conj. acid] = log(0.45/0.05) = 0.0954
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