Chapter 14: Problem 99

Calculate the $\mathrm{pH}$ of a $0.20-M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}$ solution $\left(K_{\mathrm{b}}=\right.5.6 \times 10^{-4} )$

Short Answer

Expert verified

The pH of a 0.20 M ethylamine solution can be calculated using the given $K_b= 5.6 \times 10^{-4}$. First, set up an ICE table and the Kb expression, then solve for the concentration of OH- ions, which is approximately 0.0167 M. Calculate the pOH, which is approximately 1.78, and finally, find the pH using the equation pH + pOH = 14. The pH of the solution is approximately 12.22.

Step by step solution

Write the balanced equation for the dissociation of ethylamine in water

Initially, we need to write the balanced chemical equation. Ethylamine (C2H5NH2) reacts with water (H2O) to produce ethylammonium ions (C2H5NH3+) and hydroxide ions (OH-). The equation is: \[C2H5NH2 + H2O \longleftrightarrow C2H5NH3^+ + OH^-\]

Set up the ICE table and the Kb expression

Next, we set up an ICE (Initial, Change, Equilibrium) table using the initial concentration of ethylamine given, which is 0.20 M. We can then use the table to write the expression for Kb: - At the start of the reaction: [C2H5NH2] = 0.20 M, [C2H5NH3+] = 0 M, [OH-] = 0 M - At equilibrium: [C2H5NH2] = 0.20 - x, [C2H5NH3+] = x, [OH-] = x The expression for Kb is: \[K_b = \frac{[C2H5NH3^+][OH^-]}{[C2H5NH2]}\]

Solve for x, which is the concentration of OH-.

Now we need to solve for x by plugging in the given Kb and the equilibrium concentrations from the ICE table: $5.6 \times 10^{-4} = \frac{x^2}{0.20 - x}$ To solve this equation, we can use the approximation $0.20 - x \approx 0.20$, since x is expected to be very small compared to the initial concentration 0.20 M. This will simplify the equation to: $5.6 \times 10^{-4} \approx \frac{x^2}{0.20}$ $x \approx \sqrt{(5.6 \times 10^{-4})(0.20)}$ $x \approx 0.0167$ The concentration of hydroxide ions is approximately equal to the value of x, which is 0.0167 M.

Calculate the pOH and pH of the solution

Now that we have the concentration of OH-, we can calculate the pOH and pH of the solution. To find the pOH, use the formula: \[pOH = -\log{[OH^{-}]}\] So, \[pOH = -\log{(0.0167)} \approx 1.78\] Finally, to find the pH, we use the equation: \[pH + pOH = 14\] Thus, \[pH = 14 - pOH \approx 12.22\] The pH of a 0.20 M ethylamine solution is approximately 12.22.

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Calculate the \(\mathrm{pH}\) of a $0.20-M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\( solution \)\left(K_{\mathrm{b}}=\right.\( \)5.6 \times 10^{-4} )$

Short Answer

Step by step solution

Write the balanced equation for the dissociation of ethylamine in water

Set up the ICE table and the Kb expression

Solve for x, which is the concentration of OH-.

Calculate the pOH and pH of the solution

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