A solution of paranitrophenol indicator of concentration
6 × 10 –4 (M) was prepared and indicator solution was
used in the spectrophotometric measurement. The
experimental results are tabulated as below
Condition | Form of Indicator | Absorbance |
Strongly Acidic | HIn | 0.142 |
Strongly Alkaline | In– | 0.943 |
pH = 8 | HIn + In– | 0.527 |
Given : log 2 = 0.3 ; log 3 = 0.48 ;
log 5 = 0.7 ; log 7 = 0.85
(i)What is the molar concentration of In – at pH = 8 ?
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i)
Sol.Let molar absorbance of HIn = α H In and molar absorbance of In – = 
∴ 6 × 10 –4 × α H In = 0.142 or α H In = 
and 6 × 10 –4 × 
= 0.943
or 
= 
At pH = 8, [In – ] + [H In] = 6 × 10 –4 and 
× [H In] + 
[In
– ] = 0.527
or 0.142 [H In] + 0.943 [In – ] = 0.527 × 6 × 10 –4
and 0.142 [H In] + 0.142 [In – ]
= 6 × 10 –4 × 0.142
∴ 0.801 [In – ] = 0.385 × 6 × 10 –4
or [In – ] = 
× 6 × 10 –4 = 0.48 × 6 × 10 –4
= 2.88 × 10 –4 (M)
(ii)
Sol. [H In] = (6 – 2.88) × 10 –4 = 3.12 × 10 –4 H In 
H
+ + In – K in = 
= 
= 
×10
–8 = 9.25 × 10 –9
(iii)
Sol. For acid base titration the pH at the equivalence point must lie within pK in ± 1 for a suitable indicators.
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