G.245. A 25.00 mL solution of 0.03040 M Na₂CO3 is titrated with 0.02720 M Ba(NO3)₂. Calculate pBa²+ following the addition ofthe given volumes of Ba(NO3)2. The Ksp for BaCO3 is 5.0 × 10-⁹.17.10 mL pBa²+= 6.15Ve pBa²+ = 4.182+35.90 mL pBa²+-0.0239Incorrect

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A 25.00 mL solution of 0.03040. Na₂CO3 is titrated with 0.02720 M Ba(NO3)₂. Calculate pBa- following the addition of the given volumes of Ba(NO3)₂. The Ksp for BaCO3 is 5.0 x 10-⁹. 17.10 mL pBa²+ = 6.15 Ve pBa²+ = 4.18 35.90 mL pBa²+ -0.0239 Incorrect

A 25.00 mL solution of 0.03040. Na₂CO3 is titrated with 0.02720 M Ba(NO3)₂. Calculate pBa- following the addition of the given volumes of Ba(NO3)₂. The Ksp for BaCO3 is 5.0 x 10-⁹. 17.10 mL pBa²+ = 6.15 Ve pBa²+ = 4.18 35.90 mL pBa²+ -0.0239 Incorrect

Chemistry & Chemical Reactivity

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Chapter4: Stoichiometry: Quantitative Information About Chemical Reactions

Section4.9: Spectrophotometry

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Question

G.245.

A 25.00 mL solution of 0.03040 M Na₂CO3 is titrated with 0.02720 M Ba(NO3)₂. Calculate pBa²+ following the addition of
the given volumes of Ba(NO3)2. The Ksp for BaCO3 is 5.0 × 10-⁹.
17.10 mL pBa²+
= 6.15
Ve pBa²+ = 4.18
2+
35.90 mL pBa²+
-0.0239
Incorrect

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