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Question: calculate ph of 10^-3 M sodium phenoxide Ka for phenol is 0.6*10^-10 HOW DID YOU UNDERSTAND IT WAS S...

calculate ph of 10^-3 M sodium phenoxide Ka for phenol is 0.6*10^-10 HOW DID YOU UNDERSTAND IT WAS SALT HYDROLYSIS

Answer

10.52

Explanation

Solution

Understanding Salt Hydrolysis:

Sodium phenoxide (NaOC6H5NaOC_6H_5) is a salt. To determine if it undergoes hydrolysis, we identify the acid and base from which it is formed:

  1. Cation (Na+Na^+): Comes from Sodium Hydroxide (NaOH), which is a strong base.
  2. Anion (C6H5O−C_6H_5O^-): Comes from Phenol (C6H5OHC_6H_5OH), which is a weak acid (indicated by its given KaK_a value of 0.6×10−100.6 \times 10^{-10}, which is significantly less than 1).

Since sodium phenoxide is a salt of a weak acid and a strong base, its anion (C6H5O−C_6H_5O^-) will react with water (hydrolyze) to produce hydroxide ions (OH−OH^-), making the solution basic. This process is called salt hydrolysis.

Calculation of pH:

  1. Hydrolysis Reaction: The phenoxide ion (C6H5O−C_6H_5O^-) reacts with water as follows: C6H5O−(aq)+H2O(l)⇌C6H5OH(aq)+OH−(aq)C_6H_5O^-(aq) + H_2O(l) \rightleftharpoons C_6H_5OH(aq) + OH^-(aq)

  2. Hydrolysis Constant (KhK_h): For a salt of a weak acid and strong base, the hydrolysis constant (KhK_h) is related to the ionic product of water (KwK_w) and the dissociation constant of the weak acid (KaK_a) by the formula: Kh=KwKaK_h = \frac{K_w}{K_a} Given: Kw=1.0×10−14K_w = 1.0 \times 10^{-14} (at 25°C), Ka=0.6×10−10K_a = 0.6 \times 10^{-10} Kh=1.0×10−140.6×10−10=1.666...×10−4K_h = \frac{1.0 \times 10^{-14}}{0.6 \times 10^{-10}} = 1.666... \times 10^{-4}

  3. ICE Table for Hydrolysis: Let C be the initial concentration of sodium phenoxide, C=10−3C = 10^{-3} M. Let 'x' be the concentration of OH−OH^- produced at equilibrium.

    SpeciesInitial (M)Change (M)Equilibrium (M)
    C6H5O−C_6H_5O^-10−310^{-3}−x-x10−3−x10^{-3} - x
    C6H5OHC_6H_5OH00+x+xxx
    OH−OH^-00+x+xxx

  4. Equilibrium Expression: Kh=[C6H5OH][OH−][C6H5O−]K_h = \frac{[C_6H_5OH][OH^-]}{[C_6H_5O^-]} 1.666×10−4=x⋅x10−3−x=x210−3−x1.666 \times 10^{-4} = \frac{x \cdot x}{10^{-3} - x} = \frac{x^2}{10^{-3} - x}

  5. Solving for x (Concentration of OH−OH^-): Rearrange the equation into a quadratic form: x2=1.666×10−4(10−3−x)x^2 = 1.666 \times 10^{-4} (10^{-3} - x) x2=1.666×10−7−1.666×10−4xx^2 = 1.666 \times 10^{-7} - 1.666 \times 10^{-4} x x2+1.666×10−4x−1.666×10−7=0x^2 + 1.666 \times 10^{-4} x - 1.666 \times 10^{-7} = 0

    Using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}: Here, a=1a=1, b=1.666×10−4b=1.666 \times 10^{-4}, c=−1.666×10−7c=-1.666 \times 10^{-7} x=−(1.666×10−4)±(1.666×10−4)2−4(1)(−1.666×10−7)2(1)x = \frac{-(1.666 \times 10^{-4}) \pm \sqrt{(1.666 \times 10^{-4})^2 - 4(1)(-1.666 \times 10^{-7})}}{2(1)} x=−1.666×10−4±2.775556×10−8+6.664×10−72x = \frac{-1.666 \times 10^{-4} \pm \sqrt{2.775556 \times 10^{-8} + 6.664 \times 10^{-7}}}{2} x=−1.666×10−4±6.9415556×10−72x = \frac{-1.666 \times 10^{-4} \pm \sqrt{6.9415556 \times 10^{-7}}}{2} x=−1.666×10−4±8.3316×10−42x = \frac{-1.666 \times 10^{-4} \pm 8.3316 \times 10^{-4}}{2}

    Since concentration (x) must be positive, we take the positive root: x=−1.666×10−4+8.3316×10−42=6.6656×10−42x = \frac{-1.666 \times 10^{-4} + 8.3316 \times 10^{-4}}{2} = \frac{6.6656 \times 10^{-4}}{2} x=3.3328×10−4x = 3.3328 \times 10^{-4} M

    Therefore, [OH−]=3.3328×10−4[OH^-] = 3.3328 \times 10^{-4} M.

  6. Calculate pOH: pOH=−log⁡[OH−]pOH = -\log[OH^-] pOH=−log⁡(3.3328×10−4)pOH = -\log(3.3328 \times 10^{-4}) pOH=4−log⁡(3.3328)pOH = 4 - \log(3.3328) pOH≈4−0.5228=3.4772pOH \approx 4 - 0.5228 = 3.4772

  7. Calculate pH: At 25°C, pH+pOH=14pH + pOH = 14 pH=14−pOHpH = 14 - pOH pH=14−3.4772pH = 14 - 3.4772 pH=10.5228pH = 10.5228

Rounding to two decimal places, the pH is 10.52.