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larisa86 [58]
2 years ago
14

Be sure to answer all parts. Write a balanced equation and Kb expression for the following Brønsted-Lowry base in water: benzoat

e ion, C6H5COO−. Include the states of all reactants and products in your equation. You do not need to include states in the equilibrium expression. Balanced equation: ⇌ Kb expression:
Chemistry
1 answer:
Cerrena [4.2K]2 years ago
3 0

Answer:

The balanced reaction is:-

C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}

K_b expression is:-

K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}

Explanation:

Benzoate ion is the conjugate base of the benzoic acid. It is a Bronsted-Lowry base and the dissociation of benzoate ion can be shown as:-

C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}

The expression for dissociation constant of benzoate ion is:

K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}

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Answer:

1.822 g of magnesium hydroxide would be produced.

Explanation:

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         NaOH                                           39.997

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So, 2.50 g of NaOH = \frac{2.50}{39.997} mol of NaOH = 0.0625 mol of NaOH

      4.30 g of MgCl_{2}  = \frac{4.30}{95.211} mol of MgCl_{2} = 0.0452 mol of MgCl_{2}

According to balanced equation-

2 mol of NaOH produce 1 mol of Mg(OH)_{2}    

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1 mol of MgCl_{2} produces 1 mol of Mg(OH)_{2}

So, 0.0452 mol of MgCl_{2} produce 0.0452 mol of Mg(OH)_{2}

As least number of moles of Mg(OH)_{2} are produced from NaOH therefore NaOH is the limiting reagent.

So, amount of Mg(OH)_{2} would be produced = 0.03125 mol

                                                                           = (0.03125\times 58.3197) g

                                                                           = 1.822 g

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What is the volume (in ml) of a 12.9 g piece of metal with a density of 7.25 g/cm3?
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