Question

If the Ksp of NaCl is experimentally determined to be 43.9, then what is the concentration of Na (in M) when it begins to crystallize out of solution

Answer 1

Answer:

6.63 M

Explanation:

NaCl(s) ---> Na^+(aq) + Cl^-(aq)

Given that [Na^+] = [Cl^-] = s

Where s= concentration of the both ions

Ksp = s^2

s= √Ksp

s= √43.9

s= 6.63 M

Answer 2

The concentration of Na (in M) obtained when it begins to crystallize out of solution is 6.63 M

What is solubility of product?

The solubility of product (Ksp) is defined as the concentration of products raised to their coefficient coefficients. This is illustrated below:

mX <=> nY + eZ

Ksp = [Y]^n × [Z]^e

Dissociation equation

NaCl(aq) → Na⁺(aq) + Cl¯(aq)

• Let the concentration of Na⁺ be y
• Let the concentration of Cl¯ be y

How to determine the concentration of Na⁺

• Solubility of product (Ksp) = 43.9
• Concentration of Cl¯ = y
• Concentration of Na⁺ = y =?

Ksp = [Na⁺] × [Cl¯]

43.9 =  y × y

43.9 = y²

Take the square root of both side

y = √43.9

y = 6.63 M

Thus, the concentration of Na⁺ is 6.63 M

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