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Setler79 [48]
3 years ago
10

9. Based on your work with benzoic acid, show the reaction between salicylic acid and potassium hydroxide. Label the acid, base,

conjugate acid and conjugate base.
Chemistry
1 answer:
shepuryov [24]3 years ago
3 0

Answer:

Explanation:

Salicylic Acid:

Salicylic Acid is an hydroxy acid that is found as a natural compound in plants.  It's IUPAC name is 2-hydroxybenzoic acid. Salicylic acid has an odorless white to light tan solid color. It sinks and mixes very slow with water.

Acid:  An acid is a substance that produce hydrogen ion or proton when dissolved in water

Base: A base is a substance that will neutralize an acid to yield salt and  water

Conjugate Base: This is a substance formed when an acid loses an hydrogen ion or proton when it dissolved in water.

Conjugate Acid: This is a substance formed when a base accept a proton from from any acid, when it dissolved in water.

Reaction between salicylic acid and potassium hydroxide

HOC₆H₄COOH(aq) + KOH(aq) ⇄ HOC₆H₄COOK(aq) + H₂O(l)

Acid ⇒ HOC₆H₄COOH (salicylic acid)

Base ⇒ KOH (potassium hydroxide)

Conjugate acid ⇒ H₂O (water)

conjugate base ⇒ HOC₆H₄COOK (  2-hydroxypotasium benzoate)

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The solubility product of calcium fluoride (CaF2(s); fluorite) is 310-11 at 25C. Could a fluoride concentration of 1.0 mg L-1
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The given question is incomplete. The complete question is as follows.

The solubility product of calcium fluoride () is  at 25 degrees C. Will a fluoride concentration of 1.0 mg/L be soluble in a water containing 200 mg/L of calcium?

Explanation:

Reaction equation for the given chemical reaction is as follows.

       CaF_{2} \rightleftharpoons Ca^{2+} + 2F^{-}

Therefore, expression for K_{sp} will be as follows.

        K_{sp} = [Ca^{2+}][F^{-}]^{2}

                     =

Also, moles of  per liter = \frac{\text{mass of F^{-} per L}}{\text{molar mass of F}}[/tex]

                = \frac{1.0 \times 10^{-3}}{19.0}

               = 5.263 \times 10^{-5} mol

Hence,    [F^{-}] = \frac{\text{moles of F^{-}}{volume}

                       = \frac{5.263 \times 10^{-5}}{1}

                      =  M

Now, moles of  per L = \frac{\text{mass of Ca^{2+} per L}}{\text{molar mass of Ca}}[/tex]

            = \frac{200 \times 10^{-3}}{40.1}

           =  M

Also,   [Ca^{2+}] = \frac{moles of Ca^{2+}}{volume}

                      = \frac{4.988 \times 10^{-3}}{1}

                     =  M

Hence, ionic product =

                 = (4.988 \times 10^{-3}) \times (5.263 \times 10^{-5})^{2}

                = 1.38 \times 10^{-11}

As, the ionic product is less than the K_{sp}, this means that the fluoride will be soluble in water containing the calcium.

4 0
3 years ago
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