A 1:1 salt has Ksp = 9.0 × 10⁻¹². Compute its molar solubility in pure water, in units of 10⁻⁶ M.

Practice question · Numerical answer

A 1:1 salt has $K_{sp} = 9.0 \times 10^{-12}$ . Compute its molar solubility in pure water, in units of $10^{-6}$ M.

Hints

For a 1:1 salt the constant is the square of the solubility.
Take the square root of 9.0e-12.

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Why

$s = \sqrt{9.0 \times 10^{-12}} = 3.0 \times 10^{-6}$ M. Note how a constant of order $10^{-12}$ corresponds to a solubility of order $10^{-6}$ , the square root halves the exponent.

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A 1:1 salt has Ksp=9.0×10−12K_{sp} = 9.0 \times 10^{-12}Ksp​=9.0×10−12. Compute its molar solubility in pure water, in units of 10−610^{-6}10−6 M.

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A 1:1 salt has $K_{sp} = 9.0 \times 10^{-12}$ . Compute its molar solubility in pure water, in units of $10^{-6}$ M.