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Q6. A student mixes 50.0 mL of 1.00 M Ba(OH)2 with 86.4 mL of 0.494 M H2SO4. Calculate the
mass of solid BaSO4 formed and the pH of the mixed solution.

Respuesta :

Histrionicus

The  mass of solid BaSO4 formed - 7.64 grams. and the pH of the mixed solution - 13.13

We can calculate the mass of the Ba(OH)2 by calculating the moles each solution contains

  • volume of Ba(OH)2 -

n(Ba(OH)2) = 1.00M x 0.05L

= 0.05 moles

  • The volume of H2SO4 is :

0.494M x 0.0864L

= 0.04268 moles

According to balanced reaction 1 mole Ba(OH)2 react with 1 mole H2SO4 molar ratio between Ba(OH)2 to H2SO4 is 1:1 therefore to react with 0.05 mole Ba(OH)2 required H2SO4 = 0.05 mole but H2SO4 therefore H2SO4 is limiting reactant

Ba(OH)2 + H2SO4 ----> BaSO4 + 2 H2O  

0.04268<--- 0.04268  

  • The mixed solution is Ba(OH)2 with 0.05 - 0.04268

= 0.0172 mole

  • The concentration of mixed solution is :

0.0172:(0.05 + 0.0664)

= 0.1478 M

  • The pH of mixed solution is:

Ba(OH)2 is strong base therefore dissociate completely and 1 mole of Ba(OH)2 form to mole of OH- ion therefore concentration of OH- is double than Ba(OH)2

14 - -log[0.1478]

= 14 - 0.83

= 13.13 pH

  • the mass of BaSO4 is

0.0328 x ( 137 + 32 + 16 x 4 )

= 7.64 grams.

Thus, The  mass of solid BaSO4 formed - 7.64 grams. and the pH of the mixed solution - 13.13

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