At what pH will an indicator with K_b = 1 tim | vedclass

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(B) For a basic indicator,the equilibrium is represented as: $InOH \rightleftharpoons In^+ + OH^-$. The dissociation constant is $K_b = \frac{[In^+][O

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$3$

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(B) For a basic indicator,the equilibrium is represented as: $InOH ightleftharpoons In^+ + OH^-$. The dissociation constant is $K_b = \frac{[In^+][OH^-]}{[InOH]}$. The color change occurs when $[In^+] = [InOH]$,which implies $K_b = [OH^-]$. Given $K_b = 1 	imes 10^{-11}$,we have $[OH^-] = 1 	imes 10^{-11} \ M$. Now,$pOH = -\log[OH^-] = -\log(1 	imes 10^{-11}) = 11$. Since $pH + pOH = 14$,we get $pH = 14 - 11 = 3$.

At what $pH$ will an indicator with $K_b = 1 \times 10^{-11}$ show a color change in a $1 \times 10^{-4} \ M$ solution?

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