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Shkiper50 [21]
3 years ago
8

Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory as a primary standard. It has the unw

ieldy formula of KHC8H4O4. This is often written in shorthand notation as KHP. If 34.10 mL of a barium hydroxide solution are needed to neutralize 2.050 grams of KHP, what is the concentration (mol/L) of the barium hydroxide solution
Chemistry
1 answer:
Sauron [17]3 years ago
4 0

Answer:

0.1472 mol/L is the concentration of the barium hydroxide solution.

Explanation:

2KHC_8H_4O_4+Ba(OH)_2\rightarrow Ba(KC_8H_4O_4)_2+2H_2O

Mass of potassium hydrogen phthalate = 2.050 g

Molar mass of potassium hydrogen phthalate = \frac{2.050 g}{204.2 g/mol}=0.01004 mol

According to reaction , 2 moles of potassium hydrogen phthalate reacts with 1 mole of barium hydroxide, then 0.01004 moles of potassium hydrogen phthalate will :

\frac{1}{2}\times 0.01004 mol=0.005020 mol of barium hydroxide

Moles of barium hydroxide = 0.005020 mol

Volume of the barium hydroxide solution = 34.10 mL = 0.03410 L

1 mL = 0.001 L

Molarity=\frac{Moles}{Volume(L)}

Molarity of the barium hydroxiude silution :

=\frac{0.005020 mol}{0.03410 L}=0.1472 mol/L

0.1472 mol/L is the concentration of the barium hydroxide solution.

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Explanation:

Cations (positively charged ions) can only form ionic bonds with anions (negatively charged ions). However, you can't just simply put one cation and one anion together to form a compound. Each compound needs to been neutral, or have an overall charge of 0. When cations and anions do not have charges that perfectly cancel, you need to modify the amount of each ion in the compound.

1.) Fe(CN)₂
-----> Fe²⁺ and CN⁻

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2.) FeCO₃

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3.)  Pb(CN)₄

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5 0
2 years ago
Draw the bridged bromonium ion that is formed as an intermediate during the bromination of this alkene. include hydrogen atoms,
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<h2>Answer</h2>

Bromination:

Any reaction or process in which bromine (and no other elements) are introduced into a molecule.

Bromonium Ion:

The bromonium ion is formed when alkenes react with bromine. When the π cloud of the alkene (acting as a nucleophile) approaches the bromine molecule (acting as an electrophile), the σ-bond electrons of Br2 are pushed away, resulting in the departure of the bromide anion.(2)

Mechanism:

Step 1:

In the first step of the reaction, a bromine molecule approaches the electron-rich alkene carbon–carbon double bond. The bromine atom closer to the bond takes on a partial positive charge as its electrons are repelled by the electrons of the double bond. The atom is electrophilic at this time and is attacked by the pi electrons of the alkene [carbon–carbon double bond]. It forms for an instant a single sigma bond to both of the carbon atoms involved (2). The bonding of bromine is special in this intermediate, due to its relatively large size compared to carbon, the bromide ion is capable of interacting with both carbons which once shared the π-bond, making a three-membered ring. The bromide ion acquires a positive formal charge. At this moment the halogen ion is called a "bromonium ion".

Step 2:

When the first bromine atom attacks the carbon–carbon π-bond, it leaves behind one of its electrons with the other bromine that it was bonded to in Br2. That other atom is now a negative bromide anion and is attracted to the slight positive charge on the carbon atoms. It is blocked from nucleophilic attack on one side of the carbon chain by the first bromine atom and can only attack from the other side. As it attacks and forms a bond with one of the carbons, the bond between the first bromine atom and the other carbon atoms breaks, leaving each carbon atom with a halogen substituent.

In this way the two halogens add in an anti addition fashion, and when the alkene is part of a cycle the dibromide adopts the trans configuration.

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