## Be sure to answer all parts. find the molar solubility of bacro4 (ksp= 2.1 x 10−10) in (a) pure water x 10 m (b) 1.6 x 10−3 m na

Question

Be sure to answer all parts. find the molar solubility of bacro4 (ksp= 2.1 x 10−10) in (a) pure water x 10 m (b) 1.6 x 10−3 m na2cro4 x 10 m

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1. The molar solubility of BaCrOu2084 in pure water is 1.4 x 10u207bu2075 M, while the molar solubility in 1.6 x 10u207bu00b3 M Nau2082CrOu2084 is 1.3 x 10u207bu2077 M.

Let’s consider the solution of BaCrOu2084.

BaCrOu2084(s) u21c4 Bau00b2u207a(aq) + CrOu2084u00b2u207b(aq)

To find the molar solubility (S) in pure water, we need to make an ICE chart.

BaCrOu2084(s) u21c4 Bau00b2u207a(aq) + CrOu2084u00b2u207b(aq)

I 0 0

C +S +S

E S S

The solubility product constant is:

[tex]Ksp = 2.1 times 10^{-10} = [Ba^{2+} ][CrO_4^{2-} ]= S^{2} S = sqrt{2.1 times 10^{-10}} = 1.4 times 10^{-5} M[/tex]

To calculate the molar solubility of BaCrOu2084 in 1.6 x 10u207bu00b3 M Nau2082CrOu2084, we need to consider the Nau2082CrOu2084 that is a strong electrolyte. Thus, when making the ICE chart, the initial concentration of CrOu2084u00b2u207b will be 1.6 x 10u207bu00b3 M.

BaCrOu2084(s) u21c4 Bau00b2u207a(aq) + CrOu2084u00b2u207b(aq)

I 0 1.6 x 10u207bu00b3

C +S +S

E S 1.6 x 10u207bu00b3 + S

The solubility product constant is:

[tex]Ksp = 2.1 times 10^{-10} = [Ba^{2+} ][CrO_4^{2-} ]= (S)(S+1.6times 10^{-3} )[/tex]

Since 1.6 x 10u207bu00b3 >>> S, 1.6 x 10u207bu00b3 + S u2245 1.6 x 10u207bu00b3

[tex]Ksp = 2.1 times 10^{-10} = (S)(1.6times 10^{-3} )S = 1.3 times 10^{-7} M[/tex]

The molar solubility of BaCrOu2084 in pure water is 1.4 x 10u207bu2075 M, while the molar solubility in 1.6 x 10u207bu00b3 M Nau2082CrOu2084 is 1.3 x 10u207bu2077 M.