Question

100mL of 0.003 M BaCl₂ solutions is mixed with 200mL of 0.0006M H₂SO₄ solution. Predict whether a precipitate of BaSO₄ will be formed or not.

(Ksp of BaSO₄=1.1×10^(-5))

Answer 1

Answer:

A precipitate of BaSO₄ will not be formed .

Explanation:

A precipitate will be formed if the ionic product of the solution is more than the solubility product of the solution.

Let us calculate the concentration of each ions.

$[Ba^{+2}]=\frac{[BaCl_{2}][volume]}{totalvolume}=\frac{0.003X100}{200+100}=0.001M$

$[SO_{4}^{-2}]=\frac{[H_{2}SO_{4}][volume]}{totalvolume}=\frac{0.0006X200}{200+100}=0.0004M$

Ionic product = $[Ba^{+2}][SO_{4}^{-2}]=0.001X0.0004=4X10^{-6}$

As ionic product is less than the solubility product, there will be no precipitate formed.

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Answer 2

Precipitate of BaSO₄ will not be formed.

Explanation:

Moles of barium chloride = n

Concentration of of barium chloride = 0.003 M

Volume of of barium chloride = V = 100 mL = 0.100 L

$n=0.003 M times 0.100 L=0.0003 mol$

Barium chloride has 1 barium ion and two chloride ions.

Moles of barium ions = 1 × 0.0003 mol = 0.0003 mol

Moles of sulfuric acid = n'

Concentration of sulfuric acid = 0.0006 M

Volume of of barium chloride = V = 200 mL = 0.200 L

$n=0.0006 M times 0.200 L=0.00012 mol$

sulfuric acid has 1 sulfate ion and two hydrogen  ions.

Moles of sulfate ions = 1 × 0.00012 mol = 0.00012 mol

After mixing 100 mL of barium chloirde and 200 mL of sulfiric acid.

Concentration of barium ions= $[Ba^{2+}[=\frac{0.0003 mol}{0.1 L+0.2 L}=0.001 mol/L$

Concentration of sulfate ions= $[SO_4^{2-}]=\frac{0.00012 mol}{0.1 L+0.2 L}=0.0004 mol/L$

$BaSO_4\rightleftharpoons Ba^{2+}+SO_4^{2-}$

Ionic product of barium sulfate in the solution ;

$K_i=[Ba^{2+}][SO_4^{2-}]=0.001 mol/L\times (0.0004 mol/L)=4.0\times 10^{-7}$

• $K_i<K_{sp}$ ; no Precipitation)
• $K_i>K_{sp}$ ; Precipitation

Solubility product of barium sulfate = $K_{sp}=1.1\times 10^{-5}$

So, $K_i<K_{sp}$

$4.0\times 10^{-7}<1.1\times 10^{-5}$ So, no precipitation

Precipitate of BaSO₄ will not be formed.

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