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Chapter 14: Problem 100

Calculate the \(\mathrm{pH}\) of a $0.050-M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\( solution \)\left(K_{\mathrm{b}}=\right.$ \(1.3 \times 10^{-3} )\)

Short Answer

Expert verified
The pH of the \(0.050\)-M \(\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution is approximately 12.46.

Step by step solution

01

Writing the ionization equation for the base

Ethylamine, \((\mathrm{C}_{2}\mathrm{H}_{5})_{2} \mathrm{NH}\), ionizes in water as follows: \[ (\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH} + \mathrm{H}_{2}\mathrm{O} \rightleftarrows (\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH}_{2}^{+} + \mathrm{OH}^{-} \]
02

Writing the expression for Kb

The base ionization constant, Kb, is given by: \[ K_{b} = \frac{[ (\mathrm{C}_{2}\mathrm{H}_{5})_{2}\mathrm{NH}_{2}^{+}] [\mathrm{OH}^{-}]}{[(\mathrm{C}_{2}\mathrm{H}_{5})_{2} \mathrm{NH}]} \]
03

Setting up the ICE table

An ICE table helps us write the initial concentration, the change in concentration, and the equilibrium concentration for all species involved. Initial (M): \(0.050 \qquad \qquad 0 \qquad \qquad \quad 0\) Change (M): \(-x \qquad \qquad \quad +x \qquad \qquad +x\) Equilibrium (M): \(0.050 - x \qquad x \qquad \qquad x\)
04

Substituting values into the Kb expression

Substituting the values from the ICE table into the Kb expression: \[ K_{b} = \frac{x \times x}{0.050 - x} = 1.3 \times 10^{-3} \]
05

Solving for hydroxide ion concentration

Because Kb is small, we can assume that x is much smaller than 0.050, simplifying the equation: \[ x^2 \approx 1.3 \times 10^{-3} \times 0.050 \] Solving for x (concentration of \(\mathrm{OH}^-\)): \[ x \approx \sqrt{1.3 \times 10^{-3} \times 0.050} \approx 2.88 \times 10^{-2} \ \mathrm{M} \]
06

Calculating pOH

Now, we can find the pOH by taking the negative log of the hydroxide ion concentration: \[ \mathrm{pOH} = -\log [ \mathrm{OH}^- ]= -\log ( 2.88 \times 10^{-2}) \approx 1.54 \]
07

Calculating pH

Finally, we can calculate the pH using the relationship between pOH and pH: \[ \mathrm{pH} = 14 - \mathrm{pOH} = 14 - 1.54 \approx 12.46 \] The pH of the \(0.050\)-M \(\left(\mathrm{C}_{2}\mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution is approximately 12.46.

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Most popular questions from this chapter

Will the following oxides give acidic, basic, or neutral solutions when dissolved in water? Write reactions to justify your answers. a. \(\mathrm{Li}_{2} \mathrm{O}\) b. \(\mathrm{CO}_{2}\) c. \(\mathrm{SrO}\)

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