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enyata [817]
4 years ago
5

At 25 °C, an aqueous solution has an equilibrium concentration of 0.00253 M for a generic cation, A2+(aq), and 0.00506 M for a g

eneric anion, B−(aq). What is the equilibrium constant, Ksp, of the generic salt AB2(s)? Ksp=At 25 °C, an aqueous solution has an equilibrium concentration of 0.00253 M for a generic cation, A2+(aq), and 0.00506 M for a generic anion, B−(aq). What is the equilibrium constant, Ksp, of the generic salt AB2(s)? Ksp=
Chemistry
2 answers:
trasher [3.6K]4 years ago
8 0

Answer:

6.477\times 10^{-7} is the equilibrium constant, K_{sp}, of the generic salt AB_2.

Explanation:

Solubility product constant : It is defined as the product of the concentration of the ions present in a solution raised to the power by its stoichiometric coefficient in a solution of a salt. This takes place at equilibrium only. The solubility product constant is represented as, K__{sp}.

A_xB_y\rightleftharpoons xA^{y+}+yB^{x-}

K_{sp}=[A^{y+}]^x\times [B^{x-}]^y

Equilibrium concentration for a generic cation = [A^{2+}]=0.00253 M

Equilibrium concentration for a generic anion = [B^{-}]=0.00506 M

AB^2\rightleftharpoons A^{2+}+2B^-

The expression of solubility product is given as:

K_{sp}=[A^{2+}][[B^-]]^2

K_{sp}=0.00253 M\times (0.00506 M)^2=6.477\times 10^{-7}

6.477\times 10^{-7} is the equilibrium constant, K_{sp}, of the generic salt AB_2.

aleksandrvk [35]4 years ago
5 0

Answer:

The equilibrium constant Ksp of the generic salt AB2 =  6.4777 *10^-8 M

Explanation:

Step 1: The balanced equation

AB2 ⇒ A2+ + 2B-

Step 2: Given data

Concentration of A2+ = 0.00253 M

Concentration of B- = 0.00506 M

Step 3: Calculate the equilibrium constant

Equilibrium constant Ksp of [AB2] = [A2+][B-]²

Ksp = 0.00253 * 0.00506² = 6.4777 *10^-8 M

The equilibrium constant Ksp of the generic salt AB2 =  6.4777 *10^-8 M

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5 0
3 years ago
What mass in g of NaCl is required to prepare 250.0 mL of a 0.3210 M NaCl solution? mass NaCl: Answer g Describe how to prepare
Juliette [100K]

Explanation:

Molarity = 0.3210M

Volume = 250ml

Mass = ?

Molarity is defined as the number of moles of solute per liter of solution. It's unit is mol/L.

This means 0.3210 solution contains  0.3210 moles of sodium chloride in 1.0 L of solution.

The target solution has a volume of 250ml. Upon conversion to L by dividing by 1000 we have;

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1 mole = 58.44g

0.7788 mole = x

9

x = 0.7788 * 58.44 / 1

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To prepare 0.3210 M NaCl; dissolve (45.51g / 4 = 11.3775g) of Nacl in (1 L / 4 = 250ml) of water.

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