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Identify and explain the following terms: hydrolysis, Lowry-Bronsted base, conjugate base, ampholyte, diprotic acid. Also, determine the conjugate acid of HCO3-, the conjugate base of HCO3-, and the c

Chemistry

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7.1

7.1.1

The term for an ion of a salt reacting with water is hydrolysis.

7.1.2

The reaction CO32−(aq)+H2O(l)⇌HCO3−(aq)+OH−(aq)\text{CO}_3^{2-}(\text{aq}) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HCO}_3^{-}(\text{aq}) + \text{OH}^{-}(\text{aq}) is a Lowry-Bronsted base reaction because the carbonate ion (CO32−\text{CO}_3^{2-}) accepts a proton (H+\text{H}^+) from water (H2O\text{H}_2\text{O}), forming bicarbonate (HCO3−\text{HCO}_3^{-}) and hydroxide (OH−\text{OH}^{-}). According to the Lowry-Bronsted model, a base is a proton acceptor.

7.1.3

The conjugate base of water (H2O\text{H}_2\text{O}) is OH−\boxed{\text{OH}^{-}}.

7.1.4

The carbonate ion CO32−\text{CO}_3^{2-} is a weak base. For a weak base, its dissociation constant (KbK_b) must be greater than the autoionization constant of water (Kw=1×10−14K_w = 1 \times 10^{-14} at 25 °C) but significantly less than 1.
The correct option is Kb>1×10−14\boxed{K_b > 1 \times 10^{-14}}.

7.1.5

An ampholyte is a substance that can act as both an acid (proton donor) and a base (proton acceptor).

7.1.6

The conjugate acid of the hydrogen carbonate ion (HCO3−\text{HCO}_3^{-}) is H2CO3\boxed{\text{H}_2\text{CO}_3}.

7.1.7

The conjugate base of the hydrogen carbonate ion (HCO3−\text{HCO}_3^{-}) is CO32−\boxed{\text{CO}_3^{2-}}.

7.2

7.2.1

A diprotic acid is an acid that can donate two protons (H+\text{H}^+ ions) per molecule in an aqueous solution.

7.2.2

  1. Calculate the hydrogen ion concentration from the given pH.

    The pH is given as 1.3.

    [H+]=10−pH[\text{H}^+] = 10^{-\text{pH}} [H+]=10−1.3[\text{H}^+] = 10^{-1.3} [H+]=0.0501187 mol⋅dm−3[\text{H}^+] = 0.0501187 \text{ mol}\cdot\text{dm}^{-3}

  2. Calculate the hydroxide ion concentration using the ion product of water (KwK_w).

    At 25 °C, Kw=[H+][OH−]=1×10−14K_w = [\text{H}^+][\text{OH}^{-}] = 1 \times 10^{-14}.

    [OH−]=Kw[H+][\text{OH}^{-}] = \frac{K_w}{[\text{H}^+]} [OH−]=1×10−140.0501187[\text{OH}^{-}] = \frac{1 \times 10^{-14}}{0.0501187} [OH−]=1.995×10−13 mol⋅dm−3[\text{OH}^{-}] = 1.995 \times 10^{-13} \text{ mol}\cdot\text{dm}^{-3}

    The concentration of the hydroxide ions is 1.995×10−13 mol⋅dm−3\boxed{1.995 \times 10^{-13} \text{ mol}\cdot\text{dm}^{-3}}.

    7.3

    7.3.1

  3. Calculate the initial concentration of the diprotic acid.

    Since the acid is strong and diprotic, it fully dissociates, releasing two H+\text{H}^+ ions per molecule.

    [H+]=2×Cacid, initial[\text{H}^+] = 2 \times C_{\text{acid, initial}}

    From 7.2.2, [H+]=0.0501187 mol⋅dm−3[\text{H}^+] = 0.0501187 \text{ mol}\cdot\text{dm}^{-3}.

    Cacid, initial=[H+]2C_{\text{acid, initial}} = \frac{[\text{H}^+]}{2} Cacid, initial=0.05011872C_{\text{acid, initial}} = \frac{0.0501187}{2} Cacid, initial=0.02505935 mol⋅dm−3C_{\text{acid, initial}} = 0.02505935 \text{ mol}\cdot\text{dm}^{-3}

  4. Calculate the concentration of the dilute acid after dilution.

    The initial volume of acid is V1=8 cm3=0.008 dm3V_1 = 8 \text{ cm}^3 = 0.008 \text{ dm}^3.
    The final volume of the diluted solution is V2=100 cm3=0.100 dm3V_2 = 100 \text{ cm}^3 = 0.100 \text{ dm}^3.
    Using the dilution formula C1V1=C2V2C_1V_1 = C_2V_2:

    (0.02505935 mol⋅dm−3)(0.008 dm3)=Cacid, dilute(0.100 dm3)(0.02505935 \text{ mol}\cdot\text{dm}^{-3})(0.008 \text{ dm}^3) = C_{\text{acid, dilute}}(0.100 \text{ dm}^3) Cacid, dilute=(0.02505935)(0.008)0.100C_{\text{acid, dilute}} = \frac{(0.02505935)(0.008)}{0.100} Cacid, dilute=0.002004748 mol⋅dm−3C_{\text{acid, dilute}} = 0.002004748 \text{ mol}\cdot\text{dm}^{-3}

  5. Calculate the concentration of the sodium hydroxide solution using titration data.

    The volume of dilute acid used is Va=25 cm3=0.025 dm3V_a = 25 \text{ cm}^3 = 0.025 \text{ dm}^3.
    The volume of sodium hydroxide used is Vb=14.2 cm3=0.0142 dm3V_b = 14.2 \text{ cm}^3 = 0.0142 \text{ dm}^3.
    The ratio of BASE : ACID is 2:12 : 1. This means nb=2n_b = 2 and na=1n_a = 1 in the stoichiometry.
    Using the titration formula:

    CaVana=CbVbnb\frac{C_a V_a}{n_a} = \frac{C_b V_b}{n_b} (0.002004748 mol⋅dm−3)(0.025 dm3)1=Cb(0.0142 dm3)2\frac{(0.002004748 \text{ mol}\cdot\text{dm}^{-3})(0.025 \text{ dm}^3)}{1} = \frac{C_b (0.0142 \text{ dm}^3)}{2} Cb=(0.002004748)(0.025)(2)0.0142C_b = \frac{(0.002004748)(0.025)(2)}{0.0142} Cb=0.0070589 mol⋅dm−3C_b = 0.0070589 \text{ mol}\cdot\text{dm}^{-3}

    The concentration of the sodium hydroxide solution is 0.00706 mol⋅dm−3\boxed{0.00706 \text{ mol}\cdot\text{dm}^{-3}}.

    7.3.2

    Indicator B must be used in the titration.
    The titration involves a strong diprotic acid and a strong base (sodium hydroxide). The equivalence point for a strong acid-strong base titration occurs at a pH of approximately 7. Indicator B has a pH range of 6.8-7.4, which effectively covers the equivalence point of this titration.

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