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ID: 590591 • Letter: N
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ng Jump to.. 11/19/2017 12:00 AM 4 5.3/10 11/18/2017 09:12 AM Print Calculator-d] Periodic Table Gradebook Sapling Learning A mixture is prepared using 29.00 mL of a 0.0698 M weak base (pK -4.610), 29.00 mL of a 0.0524 M weak acid (pK, 3.217) and 1.00 mL of 0.000113 M Hin and then diluting to 100.0 mL, where Hin is the protonated indicator. The absorbance measured at 550 nm in a 5.000 cm cell was 0.1161. The molar absorptivity (e) values for Hin and its deprotonated form In at 550 nm are 2.26 x10 M-cmand 1.53 x 10 M om respectively What is the pH of the solution? Incorrec See the lower panel for more information on how to solve for the pH of the solution. Number PH 10.91 hat are the concentrations of In and Hin? Number Number What is the pKs for Hin?Explanation / Answer
Volume of weak base = 29.0 mL = 0.029 L
Concentration of weak base (pKa = 4.610) = 0.0698 M
Volume of weak acid = 29.0 mL = 0.029 L
Concentration of weak acid (pKa = 3.217) = 0.0524 M
Volume of indicator = 1.0 mL
Concentration of indicator = 0.000113 M
Total volume of solution = 100 mL = 0.1 L
Number of moles of weak base = 0.029 x 0.0698 = 0.0020242 moles
Number of moles of weak acid = 0.029 x 0.0524 = 0.0015196 moles
Assuming weak acid is completely neutralized by excess weak base, the remaining total number
of moles of weak base = 0.0020242 – 0.0015196 = 0.0005046 moles
Total volume of solution = 100 mL = 0.1L
Molar concentration of remaining weak base = 0.0005046/0.1 = 0.005046 M
pOH = - log [OH] = - log [0.005046] = 2.2971
pH = 14 – pOH = 14 – 2.2971 = 11.703
pH of solution = 11.703
Assuming all of HIn converted into In by excess weak base the total number of moles of In- will be
equal to that of HIn used. However molar concentration of In- will be changed as the dilution of
solution effects.
Molarity of HIn used = 0.000113 M
Volume of HIn used = 1 mL = 0.001 L
Total number of moles of HIn used = 0.000113 x 0.001 = 0.000000113 moles
Total volume of solution = 100 mL = 0.1 L
concentration of In- , [In-], in solution = 0.000000113/0.1 = 0.00000113 M = 1.13 x 10-6 M
concentration of HIn in solution can be found from its molar absorptivity value and absorption of
solution.
Molar absorptivity value of HIn = 2.26 x 104 M-1 cm-1
Absorption of solution at 550 nm = 0.1161
Beer’s law states
A = lc
Where A = absorbance
= molar absorptivity, 2.26 x 104 M-1 cm-1
l= length of cuvette, 5 cm
c = concentration of solution HIn
concentration of HIn, [HIn] = A/l = 0.1161/2.26 x 104 x 5 = 0.1161/113000 = 1.0319 x 10-6 M
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