All categories

3 years ago

Why does a solution of a weak base and its conjugate acid act as a better buffer than does a solution?

1 answer:

german3 years ago

7 0

Hello!

A solution of a weak base and its conjugate acid act as a better buffer than does a solution of a weak base alone because <span>A solution of a weak base alone has no acid present to absorb added base.</span>

If an acid (In this case HCl) is added to a buffer or a weak base solution, the following reaction happens:

HCl + A⁻ → HA + Cl⁻

In this way, the addition of acid is neutralized by the base.

If a base is added (In this case NaOH), only a solution of a weak base and its conjugate acid can react in the following way:

NaOH + HA → NaA + H₂O

So, a solution of only a weak base can resist the addition of acids but not bases, so it isn't a good buffer.

Have a nice day!

You might be interested in

Which of the following scientists came up with the first widely recognized atomic theory?. . A. John Dalton. . B. Antione Lavois

Yuri [45]

"John Dalton" is the one scientist among the following choices given in the question that <span>came up with the first widely recognized atomic theory. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that has come to your help.</span>

8 0

3 years ago

Read 2 more answers

How long is a half life for carbon 14?

babunello [35]

Answer:

half-life of 5,700 ± 40 years

Explanation:

6 0

3 years ago

Read 2 more answers

Which of these substances has the lowest pH? 0.5 M HBr, pOH = 13.5 0.05 M HCl, pOH = 12.7 0.005 M KOH, pOH = 2.3

vagabundo [1.1K]

<h3><u>Answer;</u></h3>

0.5 M HBr, pOH = 13.5 ; Has the lowest pH

<h3><u>Explanation;</u></h3>

From the question;

pH = -Log [OH]

or pH = 14 - pOH

Therefore;

For 0.5 M HBr

[H+] = 0.5 M

pH = - Log [0.5]

= 0.30

For; pOH = 13.5

pH = 14 - pOH

= 14 -13.5

= 0.5

For; 0.05 M HCl

pH = - log [H+]

[H+] = 0.05

pH = - Log [0.05]

= 1.30

For; pOH = 12.7

pH = 14 -pOH

= 14 -12.7

= 1.30

For; 0.005 M KOH,

pOH = - log [OH]

[OH-] = 0.005

pOH = - Log 0.005

= 2.30

pH = 14 - 2.30

= 11.7

For; pOH = 2.3

pH = 14 -pOH

= 14- 2.3

= 11.7

6 0

3 years ago

Read 2 more answers

A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)

rodikova [14]

Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

$\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }$

<u>Given:</u>

A mysterious white powder could be,

powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

The solute = the powder
The solvent = ethanol
The freezing point of the solvent = −114.6°C
The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

Mass of solute = 82 mg = 0.082 g
Mass of solvent = density x volume, i.e., $\boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg \ }$

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

$\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

Now we combine it with the formula of freezing point depression.

$\boxed{ \ \Delta T_f = K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

$\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }$

$\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }$

$\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }$

We get $\boxed{ \ Mr = 153.65 \ }$

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>

The molality and mole fraction of water brainly.com/question/10861444
About the mass and density of ethylene glycol as an antifreeze brainly.com/question/4053884
About the solution as a homogeneous mixture brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

7 0

3 years ago

Read 2 more answers

How many milliliters of 2.00 M H2SO4 will react with 28.0 g of NaOH?

77julia77 [94]

Answer:175⋅mL of the given sulfuric acid

Explanation:

5 0

2 years ago