?A solution was prepared by mixing 25.00 mL of 0.080 0 M aniline, 25.00 mL of 0.060 0 M sulfanilic acid, and 1.00 mL of 1.23 x10 -4 M HIn and then diluting to 100.0 mL. (HIn stands for protonated indicator.)

The absorbance measured at 550 nm in a 5.00-cm cell was 0.110. Find the concentrations of HIn and In- and pKa for HIn.

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O NH, Anilinium ion PK₁ = 4.601 HIn €325 = 2.45 x 104 M-¹ cm-1 Esso 2.26 x 10¹ M-¹ = cm 1 -O₂S -NH₂ Sulfanilic acid pK₂ = 3.232 H+ + In 1 £325 = 4.39 x 10³ M-¹ cm-1 1.53 x 104 M-¹ cm - 1 Esso O NH, Anilinium ion PK₁ = 4.601 HIn €325 = 2.45 x 104 M-¹ cm-1 Esso 2.26 x 10¹ M-¹ = cm 1 -O₂S -NH₂ Sulfanilic acid pK₂ = 3.232 H+ + In 1 £325 = 4.39 x 10³ M-¹ cm-1 1.53 x 104 M-¹ cm - 1 Esso

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