## Chemistry and Chemical Reactivity (9th Edition)

$[H_3O^+] = 1.468 \times 10^{- 6}M$ $pH = 5.833$
1. Calculate the molar mass: 12.01* 6 + 1.01* 5 + 16* 1 + 1.01* 1 = 94.12g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.195}{ 94.12}$ $n(moles) = 2.072\times 10^{- 3}$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 2.072\times 10^{- 3}}{ 0.125}$ $C(mol/L) = 0.01657$ 4. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [C_6H_5O^-] = x$ -$[C_6H_5OH] = [C_6H_5OH]_{initial} - x = 0.01657 - x$ For approximation, we consider: $[C_6H_5OH] = 0.01657M$ 5. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][C_6H_5O^-]}{ [C_6H_5OH]}$ $Ka = 1.3 \times 10^{- 10}= \frac{x * x}{ 0.01657}$ $Ka = 1.3 \times 10^{- 10}= \frac{x^2}{ 0.01657}$ $2.155 \times 10^{- 12} = x^2$ $x = 1.468 \times 10^{- 6}$ Percent ionization: $\frac{ 1.468 \times 10^{- 6}}{ 0.01657} \times 100\% = 0.008856\%$ %ionization < 5% : Right approximation. Therefore: $[H_3O^+] = [C_6H_5O^-] = x = 1.468 \times 10^{- 6}M$ And, since 'x' has a very small value (compared to the initial concentration): $[C_6H_5OH] \approx 0.01657M$ 6. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.468 \times 10^{- 6})$ $pH = 5.833$