All categories

3 years ago

Calculate the Ka of a 0.35M weak acid with a pH of 4.2.

Chemistry

1 answer:

Nadusha1986 [10]3 years ago

8 0

Answer:

Ka = 1.14x10⁻⁸

Explanation:

First we <u>calculate [H⁺] from the pH</u>:

pH = -log[H⁺]

[H⁺] = $10^{-pH}$

[H⁺] = 6.31x10⁻⁵ M

For a monoprotic weak acid, the molar concentration of H⁺ of a solution can be expressed as:

[H⁺] = √(C*Ka)

Where C is the molar concentration of the weak acid solution.

6.31x10⁻⁵ M = $\sqrt{0.35M*Ka}$

1.14x10⁻⁸ = Ka

You might be interested in

Which solution has the higher boiling point?

dedylja [7]

NaCl has higher boiling point than C₂H₆O₂ in H₂O.

<h3>What is molality?</h3>

The number of moles of solute in a solution equal to 1 kg or 1000 g of solvent is referred to as its molality.
The definition of molarity, on the other hand, is based on a certain volume of solution.
Mol/kg is a typical molality measurement unit in chemistry.

<h3>Calculation of boiling point:</h3>

When the non-volatile solute is dissolved in a solvent, the boiling point rises along with the molality (concentration) of the solute.

Given,

Mass of C₂H₆O₂ (solute) = 15.0 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of C₂H₆O₂ = 62 g/mol

Thus, the moles of solute = 15.0 g x 1.0 mol solute/62 g/mol

= 0.2419 mol

Therefore, molality(m) of C₂H₆O₂ (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2419/0.50

= 0.4838 m

Similarly,

Given ,

Mass of NaCl (solute) = 15 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of NaCl (solute) = 58.44 g/mol

Thus, the moles of NaCl (solute) = 15.0 g x 1.0 mol solute/58.44

= 0.2566 mol

Therefore, molality(m) of NaCl (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2566 mol/0.50 kg

= 0.51 m

Hence, the molality of NaCl (solute) is more than the molality of C₂H₆O₂(solute).

So, with an increase in the solute's concentration (molality), the boiling point rises.

Therefore, NaCl has higher boiling point than C₂H₆O₂ in H₂O.

Learn more about boiling point here:

brainly.com/question/24168079

#SPJ4

7 0

2 years ago

Nitrous oxide gas is a form of

zepelin [54]

Its a form of anaesthesia used by dentists

hope that helsp

6 0

3 years ago

Calculate the maximum KE and velocity of an electron from zinc by a 275nm photon

Dmitry [639]

Answer:

Kinetic energy: $6.024*10^{-20}$

Velocity: $3.64*10^{5}$

Explanation:

From the equations of the photo-electric effect,

We know:

$hv=hv_0+K.E$

Where,

1. $h$ is the Planck's constant which is $6.626*10^{-34}$

2. $v,v_0$ are the frequency of light emitted and threshold frequencies respectively.

3. $K.E$ is the kinetic energy of the electrons emitted.

By fact, we come to know that the threshold frequency of Zn is 300nm

And also $v=\frac{c}{d}$

Where ,

1. $c$ is the speed of light $=3*10^8$

2. $d$ is the wavelength.

Thus,

$\frac{hc}{d}=\frac{hc}{d_0}+K.E\\K.E=\frac{hc}{d}-\frac{hc}{d_0}\\K.E=hc*(\frac{1}{d}-\frac{1}{d_0})\\K.E=6.626*10^{-34}*3*10^8*(\frac{1}{275*10^{-9}}-\frac{1}{300*10^{-9}})\\K.E=6.024*10^{-20}kgm^2s^{-2}$