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Chapter 6: QJ-E (page 141)

Question: Histidine is a triprotic amino acid:
Find the value of the equilibrium constant for the reaction

Short Answer

Expert verified

The equilibrium constant for the reaction is 1.17脳10^-8

Step by step solution

01

Given data

Histidine is one type of amino acid. It follows the following equilibrium constant values for protonation

$K_{1} = 3 脳 10^{- 2} K_{2} = 8.5 脳 10^{- 7} K_{3} = 4.6 脳 10^{- 10}$

We need to determine equilibrium constant for the following reaction

02

Determine the equilibrium constant

be denoted by A and the equilibrium constant for the reaction be K.

The equilibrium constant for the reaction will be

$\begin{array}{rcl} [H^{+}] [O H^{-}] & = & K_{w} \\ K_{w} & = & 10^{- 14} \\ \frac{[A H^{+}] [O H^{-}]}{[A H]} & = & K \\ \frac{[A H]}{[A]} & = & K_{2} \\ \frac{[H^{+}] [A] [O H^{-}]}{[A H]} & = & \frac{K_{w}}{K_{2}} \\ \frac{[A H^{+}] [O H^{-}]}{[A H]} & = & \frac{K_{w}}{K_{2}} \\ K & = & \frac{K_{w}}{K_{2}} \\ = & \frac{10^{- 14}}{8.5 脳 10^{- 7}} \\ = & 1.17 脳 10^{- 8} \end{array}$

Therefore, the equilibrium constant for the reaction is 1.17脳10^-8

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Most popular questions from this chapter

From the following equilibrium constants, calculate the equilibrium constant for the reaction ${HO}_{2} {CCO}_{2} H 鈬� 2 H^{+} + C_{2} O_{4}^{2 -}$

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