Question

Hln is an acidic indicator $(K_{Ind}={10}^{-7})$ which dissociates in aqueous acidic solution of $30$ mL of $0.05M$ $H_{3}P{O}_4(K_1={10}^{-3},K_2={10}^{-7},K_3={10}^{-13})$.

The value of $\left[\frac{I{n}^{⊝}}{HIn}\right]$ is:

• A. $1.413\times {10}^{-5}$
• B. $1.413\times {10}^{-4}$
• C. $3.128\times {10}^{-5}$
• D. $3.128\times {10}^{-14}$

Answer 1

The correct option is A $1.413\times {10}^{-5}$

Initial concentration of phosphoric acid is $0.05M$.
The equilibrium concentrations of phosphoric acid, $H_{2}P{O}_4^{-}$ and proton are $0.05-xM$, $xM$ and $xM$ respectively.
$Ka=\frac{\left[H^{+}\right]\left[H_{2}P{O}_4^{-}\right]}{\left[H_{3}P{O}_4\right]}0.001=\frac{x^2}{0.05-x}{x}^2-0.001x+5\times {10}^{-5}x=0.007079=\left[H^{+}\right]pH=-log\left[H^{+}\right]=-log0.007079=2.15Ka={10}^{-7}pKa=-logKa=-log{10}^{-7}=7pH=pKa+log\frac{\left[In{d}^{-}\right]}{\left[HIn\right]}2.15=7+log\frac{\left[In{d}^{-}\right]}{\left[HIn\right]}log\frac{\left[In{d}^{-}\right]}{\left[HIn\right]}=-4.85\frac{\left[In{d}^{-}\right]}{\left[HIn\right]}=1.413\times {10}^{-5}$