The following reaction: HF(aq) <--> H+(aq) + F-(aq) has an equilbirum constant (K) value of 7.2 x 10-4. This means that th
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First of all, I need to know what these five salts are. Luckily, I found a similar problem from another website which is shown in the attached picture. The Ksp is the solubility product constant. It follows the formula:
Ksp = [cation concentration]ᵃ[anion concentration]ᵇ
where a and b are the subscripts of the metal and nonmetal, respectively.
For the solutions ahead, let x be the concentration of the cation.
A. The formula is PbCr₂O₄.
2.8×10⁻¹³ = [x][x]
Solving for x, x = 5.29×10⁻⁷ M
B. The formula is Co(OH)₂.
1.3×10⁻¹⁵ = [x][x]²
Solving for x, x = 1.09×10⁻⁵ M
C. The formula is CoS.
5×10⁻²² = [x][x]
Solving for x, x = 2.24×10⁻¹¹ M
D. The formula is Cr(OH)₃.
1.6×10⁻³⁰ = [x][x]³
Solving for x, x = 3.56×10⁻⁶ M
E. The formula is Ag₂S.
6×10⁻⁵¹ = [x]²[x]
Solving for x, x = 1.82×10⁻¹⁷ M
<em>Thus,the highest concentration is letter B, Cobalt (II) Hydroxide.</em>
Ksp = [cation concentration]ᵃ[anion concentration]ᵇ
where a and b are the subscripts of the metal and nonmetal, respectively.
For the solutions ahead, let x be the concentration of the cation.
A. The formula is PbCr₂O₄.
2.8×10⁻¹³ = [x][x]
Solving for x, x = 5.29×10⁻⁷ M
B. The formula is Co(OH)₂.
1.3×10⁻¹⁵ = [x][x]²
Solving for x, x = 1.09×10⁻⁵ M
C. The formula is CoS.
5×10⁻²² = [x][x]
Solving for x, x = 2.24×10⁻¹¹ M
D. The formula is Cr(OH)₃.
1.6×10⁻³⁰ = [x][x]³
Solving for x, x = 3.56×10⁻⁶ M
E. The formula is Ag₂S.
6×10⁻⁵¹ = [x]²[x]
Solving for x, x = 1.82×10⁻¹⁷ M
<em>Thus,the highest concentration is letter B, Cobalt (II) Hydroxide.</em>