Need help with this question The pH of an unknown was determined by using an ind
ID: 517975 • Letter: N
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Need help with this question
The pH of an unknown was determined by using an indicator. The indicator is yellow below pH 3.5, green from pH 4.0-4.5, and blue above pH 5.0. In solution, the indicator's equilibrium can be expressed as: HIn + H_2O rightwardsharpoonoverleftwardsharpoon HH_3O^1+ + In^1- for which K_a = [H_3O^1 +][In^1-]/[HIn] = 1.6 times 10^-5 Using the absorbance of HIn and In^1- and the concentrations of HIn and In^1-, the pH of the buffer can be determined. Use the following lab data to determine the pH of the unknown buffer. [HIn] = 1.0 times 10^4 M Flask 100.0 mL volumetric flask, 25.00mL of Indicator, 25.00 mL of Acid, dilute to mark with di-water Flask 2: 100.0 mL volumetric flask, 25.00 mL if Indicator, 25.00 ml of Base, dilute to mark with di-water Unknown: 100.0mL Volumetric Flask, 25.00mL of Indicator, 50.00mL of unknown buffer, dilute to mark with di-waterExplanation / Answer
Henderson-Hasslebalch equation can be used to find the pH of the solution containing the indicator
pH = pKa + log ( [In-] / [HIn])
and pKa = -log (ka)
pKa = -log (1.6 * 10-5 )
= 4.795
The ratio [In-]/[HIn] can be obtained by the plot from spectrophotometric measurements which are made at two wavelengths. The first wavelength ( 1 ) is where the acidic (HIn) but not the basic form (In- ) of the indicator strongly absorbs. The second wavelength ( 2 ) is chosen where the basic but not the acid form strongly absorbs.
According to the Beer's law, the absorbance at 1 and 2 are
A 1 = (HIn,1)•b•[HIn] ________(1)
A 2 = (In_ ,2)•b•[In- ] ________(2)
where A = absorbance
= molar absorptivity
and b = cell pathlength.
At any pH, the total concentration (CT) of both forms of the indicator will be constant and equals to the sum of the individual concentrations of each species (CT = [HIn] + [In-])
In solutions of low pH, essentially all of the indicator is in the acid form. Consequently, in highly acid solutions,
CT = [HIn],
and
A 1,acidic = (HIn,1) •b•CT ________(3)
Similarly,
In highly basic solutions essentially all of the indicator is in the basic form
CT = [In-]
A 2,basic = (In_ ,2) •b•CT ________(4)
Taking a ratio of equations (1) to (3) and (2) to (4) provides expressions for the fraction of the indicator present in each form
A 1/ A 1,acidic = [HIn]/CT ________(5)
A 2/ A 2,basic = [In- ]/CT ________(6)
The ratio [In-]/[HIn] at any pH can be obtained by dividing equation (5) by equation (6) to give the ratio
[In- ]/[HIn] = (A 2 •A 1,acidic)/ (A 1 •A 2,basic)
where
A 1 is the absorbance of the indicator in unknown solution at 1 (446 nm)
A 2 is the absorbance of the indicator in unknown solution at 2 (618 nm )
A 1, acidic is the absorbance of the indicator in highly acidic solution
A 2, basic is the absorbance of the indicator in highly basic solution
so according to the equation
pH = pKa + log ( [In-] / [HIn])
pH = 4.795 + log ([In-] / [HIn])
and [In-] / [HIn] = (A 2 •A 1,acidic)/ (A 1 •A 2,basic)
= [(0.78*0.46) / (0.56*0.99)
= 0.647
so the pH of unknown solution will be
pH = 4.795 + log ([In-] / [HIn])
= 4.795 + log (0.647)
= 4.795 + (-0.189)
= 4.606
pH of unknown buffer is 4.606