QuestionApril 23, 2025

Calculate the molar solubility (mol/L) of PbCrO_(4).K_(sp)=1.8times 10^-14 Type answer: square

Calculate the molar solubility (mol/L) of PbCrO_(4).K_(sp)=1.8times 10^-14 Type answer: square
Calculate the molar solubility (mol/L) of PbCrO_(4).K_(sp)=1.8times 10^-14 Type answer: square

Solution
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Answer

1.34 \times 10^{-7} \, mol/L Explanation 1. Write the dissociation equation PbCrO_{4} \rightarrow Pb^{2+} + CrO_{4}^{2-} 2. Express K_{sp} in terms of molar solubility Let s be the molar solubility. Then, [Pb^{2+}] = s and [CrO_{4}^{2-}] = s. So, K_{sp} = [Pb^{2+}][CrO_{4}^{2-}] = s \cdot s = s^2. 3. Solve for s s^2 = 1.8 \times 10^{-14}, thus s = \sqrt{1.8 \times 10^{-14}}. 4. Calculate the value s = \sqrt{1.8 \times 10^{-14}} \approx 1.34 \times 10^{-7}

Explanation

1. Write the dissociation equation<br /> $PbCrO_{4} \rightarrow Pb^{2+} + CrO_{4}^{2-}$<br /><br />2. Express $K_{sp}$ in terms of molar solubility<br /> Let $s$ be the molar solubility. Then, $[Pb^{2+}] = s$ and $[CrO_{4}^{2-}] = s$. So, $K_{sp} = [Pb^{2+}][CrO_{4}^{2-}] = s \cdot s = s^2$.<br /><br />3. Solve for $s$<br /> $s^2 = 1.8 \times 10^{-14}$, thus $s = \sqrt{1.8 \times 10^{-14}}$.<br /><br />4. Calculate the value<br /> $s = \sqrt{1.8 \times 10^{-14}} \approx 1.34 \times 10^{-7}$
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