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匿名 發問於 科學及數學化學 · 6 年前

F.4 Chem pls help !!!

1. Why is an alkaline salt alkaline? e.g. when K2CO3 reacts with an acid it produces CO2 as well. Isn't that alkali reacts with acid to produce salt and water only?

Similar to alkaline salt, why is acidic salt acidic?

2. please write the redox equation for the reaction between concentrated

hydrochloric acid and potassium permanganate solution.

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  • 6 年前
    最愛解答

    Q1a. Why is an alkaline salt alkaline? why is acidic salt acidic?

    Ans: An easier way to check whether a salt is acidic or alkaline is to look at the cation and anion of that salt.

    If cation is stronger alkaline while anion is weak acidic, then it is an alkaline salt.

    And vice versa for acidic salt.

    e.g. K2CO3 => K+ is a strong cation (which form KOH, a strong alkaline in water),

    while CO3(2-) is a weak anion (H2CO3 is a weak acid). So, it is an alkaline salt.

    Q1b. when K2CO3 reacts with an acid it produces CO2 as well. Isn't that alkali reacts with acid to produce salt and water only?

    Ans: An alkaline salt is a SALT. It is NOT an alkaline even though its pH is >7, which is alkaline (means alkalinity 鹼度是鹼性).

    So, the equation: ALKALI + ACID → SALT + WATER

    does not apply to alkaline SALT.

    For example:

    potassium hydroxide + nitric acid → potassium nitrate + water

    KOH + HNO3 → KNO3 + H2O

    If the alkaline salt is K2CO3, a carbonate, the applicable equation is:

    CARBONATE SALT + ACID → SALT + WATER + CARBON DIOXIDE

    where CARBON DIOXIDE is produced

    For example:

    Potassium carbonate + Nitric acid → Potassium nitrate + Water + Carbon dioxide

    K2CO3 + 2HNO3 → 2KNO3 + H2O + CO2

    Q.2. Redox equation for concentrated hydrochloric acid and potassium permanganate solution.

    Ans: Mn in KMnO4 is 7+. In acidic solution of permanganate, Mn(7+) will reduce to Mn(2+) and water.

    Half equation for the oxidizing agent (permanganate ion):

    MnO4(-) + 8H(+) + 5e- → Mn(2+) + 4H2O ... (1)

    Half equation for the reducing agent (Chloride ion):

    2Cl(-) → Cl2 + 2e- ... (2)

    Balancing the 2 ionic equations, (1) x2 and (2) x5 to eliminate electrons:

    (1)x2: 2MnO4(-) + 16H(+) + 10e- → 2Mn(2+) + 8H2O

    (2)x5: 10Cl(-) → 5Cl2 + 10e-

    ==> 2MnO4(-) + 16H(+) + 10Cl(-) → 2Mn(2+) + 8H2O + 5Cl2

    To get the overall balanced equation, with 2K(+) + 6Cl(-) ions:

    ==> 2K(+) + 2MnO4(-) + 16H(+) + 10Cl(-) + 6Cl(-) → 2KCl + 2MnCl2 + 8H2O + 5Cl2

    ==> 2KMnO4(aq) + 16HCl(aq) → 2KCl(aq) + 2MnCl2(aq) + 8H2O(l) + 5Cl2(g)

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