Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 13 - Exercises - Page 575c: 68

Answer

$[H_3O^+] = 5.107 \times 10^{- 4}M$ $[C_6H_5{CO_2}^-] = 5.107 \times 10^{- 4}M$ $[C_6H_5{CO_2}H] = 4.074 \times 10^{-3}M$ $[OH^-] = 1.958 \times 10^{- 11}M$ $pH = 3.292$

Work Step by Step

1. Calculate the molar mass: 12.01* 6 + 1.01* 5 + 12.01* 1 + 16* 2 + 1.01* 1 ) = 122.13g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.56}{ 122.13}$ $n(moles) = 4.585\times 10^{- 3}$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $ C(mol/L) = \frac{ 4.585\times 10^{- 3}}{ 1} $ $C(mol/L) = 4.585\times 10^{- 3}M$ 4. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [C_6H_5{CO_2}^-] = x$ -$[C_6H_5CO_2H] = [C_6H_5CO_2H]_{initial} - x = 4.585 \times 10^{- 3} - x$ For approximation, we consider: $[C_6H_5CO_2H] = 4.585 \times 10^{- 3}M$ 5. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][C_6H_5{CO_2}^-]}{ [C_6H_5CO_2H]}$ $Ka = 6.4 \times 10^{- 5}= \frac{x * x}{ 4.585\times 10^{- 3}}$ $Ka = 6.4 \times 10^{- 5}= \frac{x^2}{ 4.585\times 10^{- 3}}$ $ 2.935 \times 10^{- 7} = x^2$ $x = 5.417 \times 10^{- 4}$ Percent dissociation: $\frac{ 5.417 \times 10^{- 4}}{ 4.585\times 10^{- 3}} \times 100\% = 11.81\%$ %dissociation < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration: $Ka = 6.4 \times 10^{- 5}= \frac{x^2}{ 4.585 \times 10^{- 3}- x}$ $ 2.935 \times 10^{- 7} - 6.4 \times 10^{- 5}x = x^2$ $ 2.935 \times 10^{- 7} - 6.4 \times 10^{- 5}x - x^2 = 0$ $\Delta = (- 6.4 \times 10^{- 5})^2 - 4 * (-1) *( 2.935 \times 10^{- 7})$ $\Delta = 4.096 \times 10^{- 9} + 1.174 \times 10^{- 6} = 1.178 \times 10^{- 6}$ $x_1 = \frac{ - (- 6.4 \times 10^{- 5})+ \sqrt { 1.178 \times 10^{- 6}}}{2*(-1)}$ or $x_2 = \frac{ - (- 6.4 \times 10^{- 5})- \sqrt { 1.178 \times 10^{- 6}}}{2*(-1)}$ $x_1 = - 5.747 \times 10^{- 4} (Negative)$ $x_2 = 5.107 \times 10^{- 4}$ - The concentration can't be negative, so $[H_3O^+]$ = $x_2$ Therefore: $[H_3O^+] = 5.107 \times 10^{- 4}M$ $[C_6H_5{CO_2}^-] = 5.107 \times 10^{- 4}M$ $[C_6H_5{CO_2}H] = 4.585 \times 10^{-3} - 5.107 \times 10^{- 4} = 4.074 \times 10^{-3}M$ 6. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 5.107 \times 10^{- 4})$ $pH = 3.292$ 7. Calculate the hydroxide concentration: $[H_3O^+] * [OH^-] = Kw = 10^{-14}$ $ 5.107 \times 10^{- 4} * [OH^-] = 10^{-14}$ $[OH^-] = \frac{10^{-14}}{ 5.107 \times 10^{- 4}}$ $[OH^-] = 1.958 \times 10^{- 11}$
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