10 mL of a saturated solution of BaSO 4 contains 1-0 - 10-7 moles of Ba 2- The K sp of BaSO 4 is A- 1 - 10-7B- 1 - 10-14C- 1 - 10-12D- 1 - 10-10

The correct option is D [Ba^+2] in saturated solution = n/V_lit = 1.0 × 10^ 7/10/1000 = 10^ 5 mol l^ 1 BaSO_4 (s) → Ba^+2 (aq) + SO_4^ 2 (aq) K_sp = [Ba^+2][ ...

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Byju's Answer

Standard IX

Chemistry

Ways of Representing the Concentration of a Solution

10 mL of a sa...

Question

$10 m L$ of a saturated solution of $B a S O_{4}$ contains $1.0 \times 10^{- 7}$ moles of $B a^{2 +}$ . The $K_{s p}$ of $B a S O_{4}$ is

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Solution

The correct option is D

$[B a^{+ 2}]$ in saturated solution = $\frac{n}{V_{l i t}}$ = $\frac{1.0 \times 10^{- 7}}{\frac{10}{1000}}$ = $10^{- 5} m o l l^{- 1}$

$B a S O_{4} (s)$ $\to$ $B a^{+ 2} (a q) + S O_{4}^{- 2} (a q)$

$K_{s p}$ = $[B a^{+ 2}] [S O_{4}^{- 2}]$ = $(10^{- 5}) (10^{- 5})$ = $10^{- 10}$

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10mL of a saturated solution of BaSO4 contains 1.0 × 10−7 moles of Ba2+. The Ksp of BaSO4 is

The correct option is D [Ba+2] in saturated solution = nVlit = 1.0 × 10−7101000 = 10−5 moll−1 BaSO4 (s) → Ba+2 (aq) + SO−24 (aq) Ksp = [Ba+2][SO−24] = (10−5)(10−5) = 10−10

$10 m L$ of a saturated solution of $B a S O_{4}$ contains $1.0 \times 10^{- 7}$ moles of $B a^{2 +}$ . The $K_{s p}$ of $B a S O_{4}$ is