What is the pH of solution A , which contains 0.0014 M indicator? What is the pH

Question

What is the pH of solution A, which contains 0.0014 M indicator?


What is the pH of solution B, which contains an undetermined amount of indicator?


Copy the data to Excel and use "Chart" to plot a graph as illustrated above. Print the Excel plot and turn it in to your lab instructor, along with equations showing how you calculated the pH.

? (nm) ABSacid ABSbase ABSsoln A ABSsoln B 400 0.332 0.178 0.205 0.239 410 0.432 0.179 0.223 0.296 420 0.542 0.181 0.244 0.360 430 0.650 0.186 0.267 0.423 440 0.742 0.194 0.290 0.478 450 0.806 0.207 0.311 0.518 460 0.832 0.225 0.331 0.538 470 0.816 0.249 0.348 0.535 480 0.761 0.280 0.364 0.513 490 0.677 0.318 0.381 0.476 500 0.578 0.364 0.402 0.432 510 0.478 0.418 0.428 0.389 520 0.387 0.477 0.461 0.354 530 0.312 0.540 0.501 0.329 540 0.256 0.606 0.545 0.315 550 0.218 0.670 0.592 0.311 560 0.195 0.731 0.637 0.315 570 0.183 0.783 0.678 0.323 580 0.179 0.824 0.712 0.332 590 0.179 0.851 0.734 0.340 600 0.182 0.862 0.744 0.344 610 0.187 0.856 0.740 0.346 620 0.193 0.833 0.722 0.342 630 0.199 0.794 0.690 0.335 640 0.205 0.740 0.647 0.323 650 0.211 0.676 0.595 0.309

Explanation / Answer

For the indicator ,

ka =3.2*10^-8

pka=-log ka=-log (3.2*10^-8)=7.5

Also total Indicator=[In]total=0.0014M

HIn <--->In- + H+

In acidic solution,

In- +H+ ---->HIn (HIn form present)

In basic solution,

HIn + OH- --->In- +H2O (In- form present)

from graph, Abs for acidic indicator(In HCl)=0.8

Abs=C*e*l where C=concentration of HIn,e=absorptivity of HIn,l=path length of light=1 cm

So, e(HIn)=Abs/C(HIn)*l=0.8/(0.0014M)*1cm=571.428 M^-1 cm^-1

e(HIn)=571.428 M^-1 cm^-1

from graph, Abs for basic indicator(in NaOH)=0.85(maximum absorption)

e(In-)=Abs/C(HIn)*l=0.9/(0.0014M)*1cm=642.857 M^-1 cm^-1

Now for unknown solution A,maximum absorption,Abs=0.7

But solution A has both HIn and In- species,so you need to find out [In-] and [HIn] first.

Abs=0.7=C(HIn)*e(HIn)*l+C(In)*e(In)*l

or,0.7=C(HIn)(571.428 M^-1 cm^-1)*(1cm) +C(In-)*(642.857 M^-1 cm^-1)*1cm

or, 0.7=C(HIn)(571.428 M^-1 ) +C(In-)*(642.857 M^-1 )

C(HIn)+C(In)=0.0014M

So,C(HIn)=0.0014M-C(In-)

or, 0.7=(0.0014M-C(In-))(571.428 M^-1 ) +C(In-)*(642.857 M^-1 )

or, 0.7=0.00098-(C(In-))(571.428 M^-1 ) +C(In-)*(642.857 M^-1 )

or,0.69902=71.429 M^-1*C(In-)

or C(In-)=0.69902/71.429M^-1=0.00979M

So C(HIn)=0.0014-0.00979=0.00839M

for solution A,

pH=pka + log [In-]/[HIn] [henderson-hasselbach eqn]

pH=7.5+log(0.00979M /0.00839M)=7.6

pH=7.6 (sol A)

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