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2 years ago

Why is the reaction of a strong acid and a strong base classified as a neutralization reaction?

Chemistry

1 answer:

Lera25 [3.4K]2 years ago

4 0

Answer:

Answer choice C

Explanation:

Basicly for the Arrhenius and Bornsted-Lowery theories of acids & bases, acid-base reactions can be divided into 4 forms...

a. Strong Acid + Strong Base (HCl/NaOH) => pH = 7 at Eqv. Pt.

b. Weak Acid + Strong Base (HOAc/NaOH) => pH > 7 at Eqv. Pt.

c. Strong Acid + Weak Base (HCl & NH₄OH) => pH < 7 at Eqv. Pt.

d. Weak Acid + Weak Base (HOAc & NH₄OH) => pH ∝ Stronger Electrolyte

*HOAc = Acetic Acid

*NH₄OH = Ammonium Hydroxide

For each type reaction the pH at equivalence point depends upon the salt generated by the acid-base reaction. Ions of the salt, if they react with water, (hydrolysis) will shift the pH up or down depending upon which ion reacts. If there is no reaction by the salt ions then the pH will depend only upon autoionization of water which gives pH = 7. Typically Strong Acids + Strong Bases will give a pH = 7 at equivalence point because the ions of the salt will not undergo hydrolysis in water.

Example:

Strong Acid + Strong Base

HCl(aq) + NaOH(aq) => NaCl(aq) + H₂O(l)

NaCl(aq) => Na⁺(aq) + Cl⁻(aq)

Na⁺(aq) + H₂O(l) => No Rxn (theoretically NaOH, but NaOH is a strong base which prefers to remain 100% ionized in water).

Cl⁻(aq) + H₂O(l) => No Rxn (theoretically HCl, but HCl is a strong acid which prefers to remain 100% ionized in water).

Therefore, the net rxn is H⁺ + OH⁻ => H₂O & pH = 7.0

Weak Acid + Strong Base

HOAc(aq) + NaOH(aq) => NaOAc(aq) + H₂O(l)

NaOAc(aq) => Na⁺(aq) + OAc⁻(aq)

Na⁺(aq) + H₂O(l) => No Rxn

OAc⁻(aq) + H₂O(l) => HOAc(aq) + OH⁻(aq) => (Excess OH⁻ functions to increase pH>7 at eqv. pt.)

Strong Acid + Weak Base

HCl(aq) + NH₄OH(aq) => NH₄Cl(aq) + H₂O(l)

NH₄Cl(aq) => NH₄⁺(aq) + Cl⁻(aq)

Cl⁻(aq) + H₂O(l) => No Rxn

NH₄⁺(aq) + H₂O(l) => NH₄OH(aq) + H⁺(aq) => (Excess H⁺ functions to decrease pH < 7 at eqv. pt. )

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2 years ago

Complete combustion of 7.50 g of a hydrocarbon produced 23.1 g of CO2 and 10.6 g of H2O. What is the empirical formula for the h

r-ruslan [8.4K]

A CH compound is combusted to produce CO2 and H2O
CnHm + O2 -----> CO2 + H2O
Mass of CO2 = 23.1g
Mass of H2O = 10.6g
Calculate by mass of the compounds
For Carbon C, divide by molecular weight of CO2 and multiply with Carbon
molecular weight. So C in grams = 23.1 x (12.01 / 44.01) = 6.3 g C
For Hydrogen H, divide by molecular weight of H2O and multiply with Hydrogen molecular weight. So H in grams = 10.6 x (2.01 / 18.01) = 0.53 g C
= 1.18 of H
Calculate the moles for C and H
6.3 grams of C x (1 mole/12.01 g C) = 0.524 moles of C
1.18 grams of H x (1 mole/1.008 g H) = 1.17 moles of H
Divides by both mole entities with smallest
C = 0.524 / 0.524 = 1 x 4 = 4
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3 years ago

Read 2 more answers

Determine the number of significant figures in each of the following

Otrada [13]

Answer:

A. 3 Significant Figures

B. 6 Significant Figures

C. 2 Significant Figures

D. 4 Significant Figures

E. 4 Significant Figures

F. 5 Significant Figures

Explanation:

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2 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

3 years ago