TABLE 10-3 Common indicators Transition Acid range (pH) color Yellow 0.05 w 1% i
ID: 584641 • Letter: T
Question
TABLE 10-3 Common indicators Transition Acid range (pH) color Yellow 0.05 w 1% in H2O 0.1 g in 26.2 mL 0.01 M NaOH. Then add -225 mL H20 0.1 g in 21.5 mL 0.01 M NaOH. Then add 225 mL H20. 0.1 g in 26.2 mL 0.01 M NaOH. Then add -225 mL H20 0.1w 1% in H2O 0.01 wt% in H2O 0.1 wt% in H2O 0.1 wt% in H2O 0.1 g in 143 mL. 0.01 M NaOH. Then add -225 mL H20 0.02 g in 60 mL cthanol. Then add 40 mL H2O 0.1 g in 23.6 mL 0.01 M NaOH. Then add -225 mL H20 0.1 g in 18.5 mL 0.01 M NaOH. Then add -225 mL H20 0.1wt% in H2O 0.1wt% in H2O 0.1 g in 16.0 mL 0.01 M NaOH. Then add-225 mL H20 0.1 g in 28.2 mL 0.01 M NaOH. Then add -225 mL H20 0.01 g in 50 mL ethanol. Then add 50 mL H20. See above. 0.1 g in 50 mL ethanol. Then add 50 mL H20. See above. See above. 0.05 g in 50 mL ethanol. Then add 50 mL. H20. 0.04 g in 50 mL cthanol. Then add 50 mL H20. 0.01 wt% in H2O Violet Yellow Yellow Yellow Methyl violet Cresol rood Thymol blue Cresol purple Erythrosine, disodium 2.2-3.6 Mcthyl orange Congo red Ethyl orange Bromocresol green Methyl red 0.0-1.6 0.2-1.8 1.2-2.8 1.2-2.8 Orange Red 3.1-4.4 3.0-5.0 3.4-4.8 3.8-5.4 4.8-6.0 4.8-6.4 5.2-6.8 5.6-7.6 5.0-8.0 6.0-7.6 6.4-8.0 6.8-8.0 7.2-8.8 7.3-8.7 7.6-9.2 8.0-9.6 8.0-9.6 8.3-10.5 10.1-12.0 Yellow Yellow Yellow Bromocresol purple p-Nitrophenol Yellow Yellow Colorless Yellow Bluc Yellow Yellow Bromothymol blue Phenol red Neutral red Cresol red Yellow Yellow Cresol purple Thymol blue Colorless Colorless Yellow Blue Pink Bluc Orange red Alizarin yellow Nitramine 10.8-13.0 Crless Orange-brown 0.1 g in 70 mL ethanol. Then add 30 mL H20. Tropacolin O 11.1-12.7 Yellow 0.1 wt% in H20Explanation / Answer
1. Titration of sodium salt of phenylalanine with HClO4
a) reaction is shown below
b) pH calculation
when 0 ml HClO4 added
say A- represent the sodium salt of phenyl alanine
salt hydrolyzes
let x amount hydrolyzed
A- + H2O <==> HA + OH-
Kb = 1 x 10^-14/6.31 x 10^-3 = x^2/0.1
x = [OH-] = 3 x 10^-7 M
pOH = -log[OH-] = 6.4
pH = 14 - pOH = 7.6
after 2.5 ml HClO4 addition
initial salt = 0.1 M x 10 ml = 1 mmol
HClO4 added = 0.1 M x 2.5 ml = 0.25 mmol
pH = pKa1 + log(base/acid)
= 2.20 + log(0.75/0.25) = 2.68
after 5 ml HClO4 addition
initial salt = 0.1 M x 10 ml = 1 mmol
HClO4 added = 0.1 M x 5 ml = 0.5 mmol
half equivalence point
pH = pKa1 = 2.20
after 7.5 ml HClO4 addition
initial salt = 0.1 M x 10 ml = 1 mmol
HClO4 added = 0.1 M x 7.5 ml = 0.75 mmol
pH = pKa + log(base/acid)
= 2.20 + log(0.25/0.75) = 1.72
after 9 ml HClO4 addition
initial salt = 0.1 M x 10 ml = 1 mmol
HClO4 added = 0.1 M x 9 ml = 0.9 mmol
pH = pKa + log(base/acid)
= 2.20 + log(0.1/0.9) = 1.24
after 10 ml HClO4 addition
initial salt = 0.1 M x 10 ml = 1 mmol
HClO4 added = 0.1 M x 10 ml = 1 mmol
Ist equivalence point
pH = 1/2(pKa1 + pKa2) = 1/2(2.20 + 9.31) = 5.755
after 11 ml HClO4 addition
initial salt = 0.1 M x 11 ml = 1 mmol
HClO4 added = 0.1 M x 11 ml = 1.1 mmol
neutral amino acid formed = 0.9 mmol
protonated amine of amino acid = 0.1 mmol
pH = 9.31 + log(0.9/0.1) = 10.26
after 15 ml HClO4 addition
initial salt = 0.1 M x 11 ml = 1 mmol
HClO4 added = 0.1 M x 15 ml = 1.5 mmol
Second half-equivalence point
pH = pKa2 = 9.31
after 20 ml HClO4 addition
initial salt = 0.1 M x 11 ml = 1 mmol
HClO4 added = 0.1 M x 20 ml = 2 mmol
second equivalence point
all amine is protonated
[protonated amine] = 1 mmol/30 ml = 0.033 M
amine gives off its proton to water and hydrolyze
say x amount is hydrolyzed
Ka2 = 4.9 x 10^-10 = x^2/0.033
x = [H3O+] = 4.02 x 10^-6 M
pH = -log[H3O+] = 5.39
after 30 ml HClO4 addition
initial salt = 0.1 M x 11 ml = 1 mmol
HClO4 added = 0.1 M x 30 ml = 3.0 M
excess [H+] = 1 mmol/40 ml = 0.025 M
pH = -log(0.025) = 1.60
c) pH at first equivalence point = 5.75
d) pH at IInd equivalence point = 5.39
f) indicator Ist equivalence point = bromocresol green
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