Question

What is the maximum concentration of Ag⁺ that can be added to a 0.00750 M solution of Na₂CO₃ before a precipitate will form? (Ksp for Ag₂CO₃ is 8.10 × 10⁻¹²)

Answer 1

Answer:

$\large \boxed{1.64\times 10^{-5}\text{ mol/L }}$

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

                           2x      0.007 50 + x

$K_{sp} =\text{[Ag ^{+} ] ^{2} [CO _{3}^{2-} ]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag ^{+} is \large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }} }$

 

Answer 2

The amount of the sample in space is referred to as concentration, in this the Maximum concentration of $Ag^+$ is $(3.28 \times 10^{-5}\ M)$.

Concentration Calculation:

• In chemistry and related sciences, the phrase "concentration" is frequently used.
• It is a metric for determining how much of one material was mixed with the other.
• The solution's concentration is indeed the amount of solute absorbed in a given quantity of liquid or solution, following are the calculation of the concentration of $Ag^+$:

        Concentration of $Na_2CO_3 = 0.00750 M$

           $(CO_3)^{2-} = Na_2CO_3 = 0.00750\ M\\\\Ksp \ \ Ag_2CO_3 =( Ag^{+})^2 (CO_3)^{2-}\\\\8.10 \times 10^{-12} = (Ag^+)^2 \times (0.00750\ M)\\\\(Ag^+)^2 = \frac{(8.10 \times 10^{-12})}{ (0.00750\ M)}\\\\(Ag^+)^2 = 1.08 \times 10^{-9}\ M^2\\\\(Ag^+) = (1.08 \times 10^{-9}\ M^2)^{\frac{1}{2}}\\\\\[(Ag^+)\] = (3.28 \times 10^{-5}\ M)$

So, the Maximum concentration of $Ag^+$ is $(3.28 \times 10^{-5}\ M)$.

Find out the more information about the concentration here:

brainly.com/question/10725862