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AveGali [126]
3 years ago
11

Titration is a type of experiment that can be performed to investigate a neutralization reaction. The equivalence point is when

all of the acid and base is fully neutralized. A sample of 0.723 M aqueous potassium hydroxide was titrated against a standard solution of hydrochloric acid. What was the volume of the potassium hydroxide solution if 79.1 mL of 1.50 M hydrochloric acid was needed to reach the equivalence point?
Chemistry
1 answer:
patriot [66]3 years ago
8 0

Answer:

The volume of the potassium hydroxide solution was 164.1 mL

Explanation:

In the equivalence point of a titration between an acid an a base, the moles of H⁺ are equal to the moles of OH⁻.

In the case of potassium hydroxide (KOH) and hydrochloric acid (HCl), an equation that represents the equivalence point would be:

V_{KOH}*M_{KOH}=  V_{HCl}*M_{HCl}

Using that equation and the data given in the problem, we can calculate the volume of potassium hydroxide:

0.723M* V_{KOH}= 79.1mL*1.50 M\\V_{KOH}=\frac{79.1mL*1.50 M}{0.723 M} \\V_{KOH}=164.1mL

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Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane  ⇄  Isobutane

At equilibrium

1.0 M               2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M               -

After equilibrium reestablishes:

(1.50-x)M            (2.5+x)

The equilibrium expression will wriiten as:

K_c=\frac{[Isobutane]}{[Butane]}

2.5=\frac{2.5+x}{(1.50-x)}

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

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Answer:

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A piece of mental with a mass of 106 g is placed in a 50 mL graduated cylinder.The volume rises from 20 mL to 31 mL. What is the
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Answer:

9.64g/mL

Explanation:

Given parameters:

Mass of the metal  = 106g

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Unknown:

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Solution:

To find the density of the metal, we use;

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