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andreyandreev [35.5K]
2 years ago
14

H A and H B are both weak acids in water, and HA is a stronger acid than HB. Which of the following statements is correct? Selec

t one: a. B is a stronger base than A^-, which is a stronger base than H_2O, which is a stronger base than CI^-. b. B is a stronger base than A^-, which is a stronger base than CI^-, which is a stronger base than H_2O. c. None of these (A-D) is correct. d. CI^- is a stronger base than A^-, which is a stronger base than B^-, which is a stronger base than H_2O e. A^- is a stronger base than B^-, which is a stronger base than H_2O, which is a stronger base than CI^-
Chemistry
1 answer:
lubasha [3.4K]2 years ago
4 0

Answer:

B is a stronger base than A^-, which is a stronger base than H2O, which is a stronger base than CI^-

Explanation:

The general equation for each acid is:

HA(aq) + H2O(ac) ⇄ H3O+(aq) + A-(aq)

HB(aq) + H2O(ac) ⇄ H3O+(aq) + B-(aq)

When these acids dissociate into its ions in water they lose a proton (H+), so they are proton donors (acids) and H2O is the proton acceptor (base). This reaction produces a conjugate acid and a conjugate base.

Conjugate base is what remains of the acid molecule after it loses a proton:

HA = acid         A- Conjugate base

HB = acid         B- Conjugate base

A conjugate acid is formed when the proton is transferred to the base

H2O = base                H3O+ = Conjugate acid

The stronger acid will produce a weaker base. According to this, if HA is a stronger acid than HB, A- would be the weaker base (B- is the stronger base).

Compared with water, A- and B- are stronger bases because when they compete for a proton they have much greater affinity for H+ than water does and the equilibrium position will lie far to the left. (HA and HB are weak acids)

Finally Cl- is the weakest base because it comes after dissociation of HCl which is a strong acid

HCl(aq) + H2O → H3O+(aq) + Cl-(aq)

Note there is no double arrows, equilibrium lies far to the right. A strong acid yields a weak conjugate base it means one that has a low affinity for a proton.

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1. A gas has a pressure of 799.0 mm Hg at 50.0 degrees C. What is the temperature at standard Pressure?
larisa [96]

Answer:

A = 674.33mmHg

B = 0.385atm

Explanation:

Both question A and B requires the application of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided that volume is kept constant.

Mathematically,

P = kT, k = P / T

P1 / T1 = P2 / T2 = P3 / T3 =.......= Pn/Tn

A)

Data:

P1 = 799mmHg

T1 = 50°C = (50 + 273.15) = 323.15K

P2 = ?

T2 = 273.15K

P1 / T1 = P2 / T2

Solve for P2

P2 = (P1 × T2) / T1

P2 = (799 × 273.15) / 323.15

P2 = 674.37mmHg

The final pressure is 674.37mmHg

B)

P1 = 0.470atm

T1 = 60°C = (60 + 273.15)K = 333.15K

P2 = ?

T2 = 273.15K

P1 / T1 = P2 / T2

Solve for P2,

P2 = (P1 × T2) / T1

P2 = (0.470 × 273.15) / 333.15

P2 = 0.385atm

The final pressure is 0.385atm

7 0
1 year ago
Read 2 more answers
When 2.36g of a nonvolatile solute is dissolved in 100g of solvent, the largest change in freezing point will be achieved when t
nignag [31]

Answer:

Option c → Tert-butanol

Explanation:

To solve this, you have to apply the concept of colligative property. In this case, freezing point depression.

The formula is:

ΔT = Kf . m . i

When we add particles of a certain solute, temperature of freezing of a solution will be lower thant the pure solvent.

i = Van't Hoff factor (ions particles that are dissolved in the solution)

At this case, the solute is nonvolatile, so i values 1.

ΔT = Difference between  fussion T° of pure solvent - fussion T° of solution.

T° fussion paradichlorobenzene = 56 °C

T° fussion water = 0°

T° fussion tert-butanol = 25°

Water has the lowest fussion temperature and the paradichlorobenzene has the highest Kf. But the the terbutanol, has the highest Kf so this solvent will have the largest change in freezing point, when all the molalities are the same.

3 0
2 years ago
BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w
Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>

Explanation:

Given: The base dissociation constant: K_{b} = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: K_{w} = 1 × 10⁻¹⁴

<em><u>The acid dissociation constant </u></em>(K_{a})<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>

K_{a}. K_{b} = K_{w}    

\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}

\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}

<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>

Reaction involved: BH⁺  +  H₂O  ⇌  B  +  H₃O+

Initial:                     0.1 M                    x         x            

Change:                   -x                      +x       +x

Equilibrium:           0.1 - x                    x         x

<u>The acid dissociation constant: </u>K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}

\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}

\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}

As, x

\Rightarrow 0.1 - x = 0.1

\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }

\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}

\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}

<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>

5 0
1 year ago
What is the change in enthalpy in kilojoules when 3.24 g of CH3OH is completely reacted according to the following reaction 2 CH
vodka [1.7K]

Answer:

12.78 kJ

Explanation:

The correct balanced reaction would be

2CH_3OH\rightarrow 2CH_4+O_2\Delta H=252.8\ \text{kJ}

Mass of methanol = 3.24\ \text{g}

Moles of methanol can be obtained by dividing the mass of methanol with its molar mass (32.04\ \text{g/mol})

\dfrac{3.24}{32.04}=0.10112\ \text{moles}

Enthalpy change for the number of moles is given by

\dfrac{\text{Number of moles of methanol in the reaction}}{\text{Enthalpy change in the reaction}}=\dfrac{\text{Number of moles in 3.24 g of methanol}}{\text{Enthaply in change in the mass of methanol}}

\\\Rightarrow\dfrac{2}{252.8}=\dfrac{0.10112}{\Delta H}\\\Rightarrow \Delta H=\dfrac{0.10112\times 252.8}{2}\\\Rightarrow \Delta H=12.781568\approx 12.78\ \text{kJ}

The change in enthalpy is 12.78 kJ.

5 0
2 years ago
Air is transferred from a 75 L tank where the pressure is 125 psi and the temperture is 288 k to a tire with a volume of 6.1 L a
aniked [119]
<h3><u>Answer</u>;</h3>

 = 4.68 K

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

According to the combined gas law;

P1V1/T1 = P2V2/T2

Given; P1 = 125 Psi

V1 = 75 L

T1 = 288 K

P2 = 25 PSI

V2 =6.1 L

Therefore;

T2 = P2V2T1/P1V1

    = (25×6.1 ×288)/(125×75)

    = 4.6848

    = 4.68 K

4 0
2 years ago
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