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2 years ago

Do all titrations of a strong base with a strong acid have the same ph at the equivalence point?

Chemistry

1 answer:

Luba_88 [7]2 years ago

7 0

Answer:

Yes, all titrations of a strong base with a strong acid have the same pH at the equivalence point.

This pH is 7.

Explanation:

Strong acids and strong bases ionize completely in aqueous solutions. The ionization of strong acids produce hydronium ions, H₃O⁺, and the ionization of strong bases produce hydroxide ions, OH⁻.

Since the ionization of strong acids and bases progress until completion, there is not reverse reaction.

The definition of pH is pH = - log [H₃O⁺]. Acids have low pH (below 7, and greater than 0) and bases have high pH (above 7 and less than 14). Neutral solutions have pH = 7.

Acid-base titrations are a method to determine the concentration of an acid from the known concentration of a base, or the concentraion of a base from the known concentration of an acid.

The equivalence point of the titration is the point at which the the number of moles of hydronium ions and hydroxide ions are equal.

Then, at that point, the hydronium and hydroxide ions will be in the stoichiometric proportion to form a neutral solution, i.e. the pH of the solution wiill be 7.

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What are the units of density

Nataliya [291]

Answer:

Kg/ ml

Explanation:

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2 years ago

why is it much easier for group 14 elements to become stable by sharing instead of transferring electrons9

Effectus [21]

Because they are closer to the farther end of the periodic table. Since they are closer to the farther end they don't want to give away their electrons because it would be easier for them to just steal them from other atoms.

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3 years ago

Read 2 more answers

Suppose that a certain fortunate person has a net worth of $77.0 billion ($). If her stock has a good year and gains $3.20 billi

marissa [1.9K]

Answer:

New net worth to three significant figures = $80.2 billion

Dollars are given to charity = $100,250,000

Explanation:

Net worth = $77.0 billion

stock Gains = $3.20 billion

New net worth = $77.0 billion + $3.20 billion = $80.20 billion

New net worth to three significant figures = $80.2 billion

One-eighth of a percent = (1/8 x 1) / 100 = 0.00125

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3 0

3 years ago

Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

3 years ago

What orbitals are used to form the 10 sigma bonds in propane (ch3ch2ch3)? Label each atom with the appropriate hybridization. Dr

SVETLANKA909090 [29]

Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.

For the compound, $CH_3CH_2CH_3$ the electronic configuration of the atoms, carbon and hydrogen are:

Carbon (atomic number=6): In ground state= $1s^{2}2s^{2}2p^{2}$

In excited state: $1s^{2}2s^{1}2p^{3}$

Hydrogen (atomic number=1): $1s^{1}$

All the bonds in the compound is single bond( $\sigma$ -bond) that is they are formed by head on collision of the orbitals.

The structure of the compound is shown in the image.

The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.

In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in $1s^{1}$ of hydrogen will overlap to the 2p^{3}-orbital of carbon.

Thus, the hybridization of Hydrogen is $s$ -hybridization and the hybridization of Carbon is $sp^{3}$ -hybridization.

The hybridization of each atom is shown in the image.

3 0

2 years ago