1. The solubility of BaSO4 in water at 25°C is measured to be 0.0023gL. Use this

Question

1. The solubility of BaSO4 in water at 25°C is measured to be 0.0023gL. Use this information to calculate Ksp for BaSO4.

2.The pH of a 0.33 M solution of HC4H3N2O3 is measured to be2.25. Calculate the acid dissociation constant Ka of barbituric acid. Round your answer to two significant figures.

3. A solution is prepared at 25 degrees C that is initially 0.096M in chlorous acid, a weak acid with Ka=1.1x10^-2, and 0.47 M in NaClO2. Calculate the pH of the solution

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Explanation / Answer

1)

Molar mass of BaSO4 = 1*MM(Ba) + 1*MM(S) + 4*MM(O)

= 1*137.3 + 1*32.07 + 4*16.0

= 233.37 g/mol

Molar mass of BaSO4= 233.37 g/mol

s = 2.3*10^-3 g/L

To covert it to mol/L, divide it by molar mass

s = 2.3*10^-3 g/L / 233.37 g/mol

s = 9.856*10^-6 g/mol

The salt dissolves as:

BaSO4 <----> Ba2+ + SO42-

   s s

Ksp = [Ba2+][SO42-]

Ksp = (s)*(s)

Ksp = 1(s)^2

Ksp = 1(9.856*10^-6)^2

Ksp = 9.71*10^-11

Answer: 9.71*10^-11

Feel free to comment below if you have any doubts or if this answer do not work

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