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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 1.17×10^-8

Step by step solution

01

Given data

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

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

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

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

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

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Most popular questions from this chapter

Find $[{Cu}^{2 +}]$ in a solution saturated with ${Cu}_{4} {(OH)}_{6} ({so}_{4})$ if $[{OH}^{-}]$ is fixed at $1.0 \times 10^{- 6} M$ . Note that ${Cu}_{4} {(OH)}_{6} ({so}_{4})$ gives 1mol of ${SO}_{4}^{2 -}$ 4 mol for of ${Cu}^{2 +}$ .

${Cu}_{4} {(OH)}_{6} ({so}_{4}) (s) ⇋ 4 C u^{2 +} + 6 O H^{-} + S O_{4}^{2 -} K_{s p} = 2.3 \times 10^{- 69}$

Reaction 6-8 is allowed to come to equilibrium in a solution initially containing $0.0100 {MBrO}_{3}^{-}$ , $0.0100 {MCr}^{3 +}$ and 1.00MH⁺. To find the concentrations at equilibrium, we construct the table at the bottom of the page showing initial and final concentrations. We use the stoichiometry coefficients of the reaction to say that if $x m ol$ of ${Br}^{-}$ are created, then we must also make x mol of ${Cr}_{2} O_{7}^{2 -}$ and 8x mol of H⁺. To produce x mol of ${Br}^{-}$ , we must have consumed x mol of ${Br}^{-} O_{3}^{-}$ and 2x mol of Cr³⁺.

(a) Write the equilibrium constant expression that you would use to solve for x to find the concentrations at equilibrium. Do not try to solve the equation.

(b) Because $K = 1 \times 10^{11}$ , we suppose that the reaction will go nearly "to completion." That is, we expect both the concentration of ${Br}^{-}$ and ${Cr}_{2} O_{7}^{2 -}$ to be close to 0.00500M an equilibrium. (Why?) That is, $x \approx 0.00500 M$ . With this value of $x, [H^{+}] = 1.00 + 8 x = 1.04 M$ and $[{BrO}_{3}^{-}] = 0.0100 - x = 0.0050 M$ . However, we cannot say $[{Cr}^{3 +}] = 0.0100 - 2 x = 0$ , because there must be some small concentration of ${Cr}^{3 +}$ at equilibrium. Write $[{Cr}^{3 +}]$ for the concentration of ${Cr}^{3 +}$ and solve for $[{Cr}^{3 +}]$ . The limiting reagent in this example is ${Cr}^{3 +}$ . The reaction uses up before consuming ${BrO}_{3}^{-}$ .

Henry's law states that the concentration of a gas dissolved in a liquid is proportional to the pressure of the gas. This law is a consequence of the equilibrium

$X (g) \overset{K_{h}}{⇋} X (a q) K_{h} = \frac{[x]}{p_{X}}$

where $K_{h}$ is called the Henry's law constant. For the gasoline additive MTBE, $K_{h} = 1.71 M / bar$ . Suppose we have a closed container with aqueous solution and air in equilibrium. If the concentration of MTBE in the water is $1.00 x 10^{2} ppm$ $(= 100 μg MTBE / g solution \approx 100 μg / mL)$ , what is the pressure of MTBE in the air?

$C H_{3} - O - C {(C H_{3})}_{3}$ Methyl- -butyl ether (MTBE, FM 88.15)

Write the $K_{b}$ reaction for pyridine and for sodium 2-mercaptoethanol.

Use electron dot structures to show why tetra methyl ammonium hydroxide, ${({CH}_{3})}_{4} N^{+} {OH}^{-}$ , is an ionic compound. That is, show why hydroxide is not covalently bound to the rest of the molecule.

See all solutions

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