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Ugo [173]
3 years ago
12

An analytical chemist is titrating of a solution of propionic acid with a solution of . The of propionic acid is . Calculate the

pH of the acid solution after the chemist has added of the solution to it.
Chemistry
1 answer:
motikmotik3 years ago
3 0

Answer:

Heres my attempt at this and hope it helps friend

.1372L * (.83M/L) = .114 mols propanoic acid

.06862L * (1.1M/L) = 0.0755 mols NaOH

using the henderson hasselbach equation

ph = pKa + log([A-]/[HA])

so

ph = 4.89 + log(.0755/.114) = 4.72

Explanation:

Identify at least three reasons the Articles of Confederations failed as a governing document. In your opinion, evaluate which defect was most debilitating, using evidence and your knowledge of American government to justify your position.

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Answer: The number of neutrons will increase as we move from left to right in a periodic table.

Explanation:

Atomic number is equal to the number of protons.

Mass number is the sum of number of neutrons and number of protons.

As we move from left to right, both the atomic number and mass number increases.

For example: As we move from Lithium to berrylium to boron to carbon to nitrogen to oxygen to fluorine to neon , the number of neutrons increase from 4 to 5 to 6 to 6 to 7 to 8 to 10 to 10.

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Part III.The two reactions involved in quantitatively determining the amount of iodate in solution are: IO3-(aq) 5 I-(aq) 6 H (a
slega [8]

Answer:

\large \boxed{\math{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}

Explanation:

The I₂ is the common substance in the two equations.

(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O

{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻

From Equation (1), the molar ratio of iodate to iodine is

\dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{3}{1}

From Equation (2), the molar ratio of iodine to thiosulfate is

\dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} = \dfrac{2}{1}

Combining the two ratios, we get

\text{Stoichiometric factor} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{IO}_{3}^{-}} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} \times \dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{2}{1} \times \dfrac{3}{1} = \mathbf{\dfrac{6}{1}}\\\\\text{The stoichiometric factor is $\large \boxed{\mathbf{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$}

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3 years ago
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Because it can provide immediate and important information about a sample's purity and whether or not a reaction is still ongoing, TLC is a common technique in the organic chemistry lab. A TLC plate can be finished in less than 5 minutes when low polarity solvents are used.

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this is the answer

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