What is the pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid with 15.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive. K a = 1.8 ×× 10-5 for CH3CO2H.

Answer:

[tex]pH=10.45[/tex]

Explanation:

Hello,

In this case, for the dissociation of the given base, we have:

[tex]base\rightleftharpoons OH^-+CA[/tex]

Whereas CA accounts for conjugated acid and OH⁻ for the conjugated base. In such a way, equilibrium expression is:

[tex]Kb=\frac{[OH^-][CA^+]}{[base]}[/tex]

And in terms of the reaction extent [tex]x[/tex] we can write:

[tex]1.6x10^{-6}=\frac{x*x}{0.05M-x}[/tex]

For which the roots are:

[tex]x_1=-0.000284M\\x_2=0.000282M[/tex]

For which clearly the result is the positive root which also equals the concentration of hydroxyl ions and we can compute the pOH:

[tex]pOH=-log([OH^-])=-log(0.000282)\\\\pOH=3.55[/tex]

And the pH:

[tex]pH=14-pOH=14-3.55\\\\pH=10.45[/tex]

Regards.

The pH of the solution is 10.45.

Let us represent codeine with the generic formula BH. We can set up the ICE table as follows;

              :B(aq) + H2O(l) ⇄ BH(aq)  + OH^-(aq)

I            0.05                        0                0

C           -x                            +x                +x

E        0.05 - x                      x                  x

We know that the Kb of codeine is 1.6 x 10^-6, Hence;

1.6 x 10^-6 = x^2/0.05 - x

1.6 x 10^-6 (0.05 - x ) =  x^2

8 x 10^-8 - 1.6 x 10^-6x =  x^2

x^2 +  1.6 x 10^-6x - 8 x 10^-8 = 0

x = 0.00028 M

The concentration of hydroxide ions = 0.00028 M

Given that pOH = - log[0.00028 M]

pOH = 3.55

pH + pOH = 14

pH = 14 - 3.55

pH = 10.45

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Answer:

measure the parallax and use it in calculations

Explanation:

got it right on test

There are many more stars at different distances from the earth. The distance to the stars calculated in light years and it is measured using parallax method.Thus option a is correct.

Parallax method is used to measure the approximate distance of stars from earth. It uses the position of nearby star from two points opposite to earth and the small angular displacement observed from the remote stars are noted.

The orbit radius of earth and distance to the stars can be calculated from the parallactic angle p, that is one second of arc. Thus the distance is described in the units parsec.

The distance to the stars are usually calculated in light years. One parsec equals 3.26 light years. The nearest star to earth is named as proxima century having the distance parallax 0.76813'' which equals 4.24 light years. Thus, parallax is inversely proportional to the distance.

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Answer: A

Explanation: Calculate the molar hydrogen ion concentration of each of the following biological solutions given the pH, Urine pH= 4.90

Answer:

d

Explanation:answer is d on edg 2020

Answer:

OPTION C is correct

C) 1.07 g

Explanation:

CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)3(aq)

But we know molarity molarity= number of moles of solute/ volume of the solution

M= n/V

From the equation above

number of moles of Cacl2 = (37.5 ×0 .100 × 10^-3) = 0.00375 moles.

Then

1 mole of Cacl2 yields 2 moles of Agcl2

0.00375 moles of Cacl2 will produce let say Y.

Y= (0.00375 ×2)/1

= 0.0075 moles.

Number of moles of Agcl2 = mass /molar mass of Agcl

Molar mass of Agcl = 178

Then mass = 178 ×0.0075 = 1.047

Therefore, the mass of agcl precipitate is

1.07 g

Answer:

Any ionic molecule formed of a base and an acid, which dissolves in water to produce ions is known as a salt. The four common types of salts are:  

1. NaCl or sodium chloride is the most common kind of salt known. It is also known as table salt.  

2. K2Cr2O7 or potassium dichromate refers to an orange-colored salt formed of chromium, potassium, and oxygen. It is toxic to humans and is also an oxidizer, which is a fire hazard.  

3. CaCl2 or calcium chloride looks like table salt due to its white color. It is broadly used to withdraw ice from roads. It is hygroscopic.  

4. NaHSO4 or sodium bisulfate produces from hydrogen, sodium, oxygen, and sulfur. It is also known as dry acid. It has commercial applications like reducing the pH of swimming pools and spas and others.  

Answer:

The reaction is given as:

[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]

Explanation:

Bases are defined as those chemical substances which give hydroxide ions in their aqueous solutions.

[tex]BOH(s)\rightarrow B^+(aq)+OH^-(aq)[/tex]

When sodium hydroxide is added to water it gets dissociated into two ions that are sodium ions and hydroxide ions. Along with this heat energy also releases during this reaction.

The reaction is given as:

[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]

The equation to show how NaOH behaves in water is NaOH → Na⁺ + (OH)⁻

The compound that produce negative hydroxide (OH−) ions when dissolved

in water are called bases .

This compounds NaOH (sodium hydroxide) is an example of a base.

When it dissolves in water it dissociate to form  negative hydroxide (OH−)

ions  and positive sodium (Na+) ions.

It can be represented by the following equation:

NaOH → Na⁺ + (OH)⁻

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Answer:

Option D. A lipid with three unsaturated fatty acid.

Explanation:

The molecule in the diagram above contains three fatty acid.

A careful observation of the molecule reveals that each of the three fatty acids contains a double bond.

The presence of a double bond in a compound shows that the compound is unsaturated.

Thus, we can say that the molecule is a lipid with three unsaturated fatty acid.

Answer:

Explanation:

mass -  15.8 g = 0.0158 kg

volume = 32.5 - 22.5 = 10.2 ml

density = mass / volume

= 0.0158 / 10.2

= 0.00154 kg/ml

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Answer:

CO2

Explanation:

Mass of carbon = 27.29

Mass of oxygen = 72.21

Step 1:

We have to first convert these masses to moles

Carbon = 2.29/2.01 = 2.274

Oxygen = 72.71/16 = 4.544

Step 2:

We have to divide the mols by the smallest mol to get simplest ratio of whole number.

The smallest mol is 2.274. we have to divide the mols by this.

2.274/2.274 = 1

4.544/2.274 = 2

Our empirical formula is therefore CO2

Answer:

half life=0.693/rate constant =188.3

The half-life of this element is 188.32 years

The formula for calculating the half-life of an element is expressed according to the equation:

[tex]t_{1/2}=\frac{ln 2}{\lambda}[/tex]

Given the following parameters:

The rate constant for the decay = 3.68 × 10⁻³ day⁻¹.

Substitute into the formula for calculating the half-life as shown:

[tex]3.68\times 10^{-3}=\frac{0.693}{\lambda}\\ 0.00368=\frac{0.693}{\lambda}\\\lambda=\frac{0.693}{0.00368}\\\lambda = 188.32 years[/tex]

Hence the half-life of this element is 188.32 years

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Answer: The volume of liquid used will be 18.055 mL.

Answer:

13.11 g.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below :

N2 + O2 —> 2NO

Next, we shall determine the mass of O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 1 x 32 = 32 g.

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO from the balanced equation = 2 x 30 = 60 g.

From the balanced equation above,

32 g of O2 reacted to produce 60 g of NO.

Finally, we shall determine the theoretical yield of NO as follow:

From the balanced equation above,

32 g of O2 reacted to produce 60 g of NO.

Therefore, 6.99 g of O2 will react to produce = (6.99 x 60)/32 = 13.11 g of NO.

Therefore, the theoretical yield of nitrogen monoxide, NO is 13.11 g.

Answer:

[tex]6.4knips[/tex]

Explanation:

Hello,

In this case, given the stated equivalences, we can use the following proportional factor in order to compute the required knips:

[tex]knips=1.00knop*\frac{6bippy}{1knop} *\frac{8pringle}{10bippy}* \frac{3blip}{18pringle} *\frac{4clips}{5blips} *\frac{10knip}{1clip} \\\\=6.4knips[/tex]

Regards.

1 knop=6.4 knips

First convert knop to bippy:-

[tex]1\ knop\times\frac{6\ bippy}{1\ knop} =6\ bippy[/tex]

Now, Convert 6 bippy to pringle:-

[tex]6\ bippy\times\frac{8\ pringle}{10\ bippy} =4.8\ pringle[/tex]

Now, convert 4.8 pringle to blip:-

[tex]4.8\ pringle\times\frac{3\ blip}{18\ priangle} =0.8\ blip[/tex]

Now, convert 0.8 blip to clips as follows:-

[tex]0.8\ blip\times\frac{4\ clips}{5\ blip} =0.64\ clip[/tex]

Now, convert 0.64 clip to knips:-

[tex]0.64\ clip\times\frac{10\ knip}{1\ clip} =6.4\ knip[/tex]

Hence, the following conversion is as follows:-

1.00 knop=6.4 knips

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Answer:

The correct answer is: 0.064 V

Explanation:

In a oxidation-reduction reaction, the relation between Gibbs free energy (ΔGº) and cell potential (Eºcell) is given by:

[tex]\Delta G^{0} = -nFE^{0} _{cell}[/tex]

where n is the number of electrons that are transferred in the reaction and F is the Faraday constant (96,500 C/mol e-). Given: ∆G° = +18.55 kJ and n= 3 mol e-, we calculate Eºcell as follows:

+18.55 kJ = (-3 mol e-) x (96500 C/mol e-) x Eºcell

Eºcell= (+18.55 kJ)/(-3 mol e-) x (96500 C/mol e-)

Eºcell= (18550 J)/ (289500 C)

Eºcell= 0.064 J/C

Since 1 Volt= 1 Joule/1 Coulomb, thus:

Eºcell= 0.064 J/C = 0.064 V

Answer:

Bakelite is a polymer made up of the monomers phenol and formaldehyde. This phenol-formaldehyde resin is a thermosetting polymer.

Answer: The monomers of bakelite are formaldehyde and phenol

Explanation:

Answer:

2Cl——>Cl2+2e-

Explanation:

It shows an electron loss or gain

Answer:

15. Critical angle of glass and water combination, θ = 62.45°

16. Critical angle for the interface between Mystery A and glass, θ = 37.93°

Note; The question is incomplete. The complete question is as follows:

Medium Air Water Glass Mystery A Mystery B Table-2 Speed (m/s) 1.00 C 0.75 c 0.67 0.41 c 0.71 c n 1.00 1.33 1.50 Index of Refraction n of a given medium is defined as the ratio of speed of light in vacuum, c to the speed of light in a medium, v. n = c/v

Table-4: Incident Angle (degrees) Reflected Angle Refracted angle (degrees) (degrees) % Intensity of reflected ray 0 10 20 30 40 50 N/A N/A N/A 30 40 50 0 11.3 22.7 34.2 46.3 59.5 N/A N/A N/A 0.67 1.22 3.08 % Intensity of refracted ray 100 100 100 99.33 98.78 96.92

When rays travel from a denser medium to a less dense medium, we can define a critical angle of incidence θ such that refracted angle θ₂ = 90°. Applying Snell's law: Critical angle θ = sin-1(n₂/n₁).

When the angle of incidence is greater than the critical angle, 100% of the light intensity is reflected. This is called total internal reflection because all the light is reflected.

15. Calculate the critical angle of glass and water combination. Show your calculation.

16. What is the critical angle for the interface between Mystery A and glass?

Explanation:

15.  Applying Snell's law; Critical angle θ = sin-1(n₂/n₁).

where n₂,refractive index of water = 1.33, n₁, refractive index of glass = 1.50 since glass is denser than water

θ = sin-1(1.33/1.50)

θ = 62.45°

Critical angle of glass and water combination, θ = 62.45°

16.  Refractive index of mystery A , n = c/v

where v = 0.41 c

therefore, n = c / 0.41 c = 2.44

Critical angle for the interface between Mystery A and glass, θ = sin-1(n₂/n₁).

where n₂,refractive index of glass = 1.50, n₁, refractive index of mystery A = 2.44 since mystery A is denser than glass as seen from its refractive index

θ = sin-1(1.50/2.44)

θ = 37.93°

Critical angle for the interface between Mystery A and glass, θ = 37.93°

Answer:

It would increase by a factor of 2

Explanation:

The rate law for a fist order reaction is given as;

A --> B

rate = k [A]

upon doubling the concentration, we have;

rate = k [A]

rate 2 = k 2 [A]

Dividing both equations;

rate 2 / rate 1 = k 2 [A] / k [A]

rate 2 / rate 1  = 2 / 1

The ratio between rate 2 and rate 1 is 2 : 1. This means that the reaction rate would also increase by a factor of 2.

Answer:

D. Rare

Explanation:

Qualitative research has to do with non-numerical data and is used to understand the beliefs of a group of people which can be gotten from surveys, questionnaires, interviews, etc.

A qualitative researcher may select a RARE case that is unusually rich in information pertaining to the research question.

This is because he wants to get an insight into the why, how, where and when of that particular ase as it's not a usual occurrence.

Answer:

The best reducing agent is Zn(s)

Explanation:

A reducing agent must to be able to reduce another compound, by oxidizing itself. Consequently, the oxidation potential must be high. The oxidation potential of a compound is the reduction potential of the same compound  with the opposite charge. Given the reduction potentials, the best reducing agent will be the compound with the most negative reduction potential. Among the following reduction potentials:

Cd₂⁺(aq)|Cd(s) ⇒ -0.40 V

Zn²⁺(aq)|Zn(s) ⇒ -0.76 V

Ni²⁺(aq)|Ni(s) ⇒‑0.25 V

Cu²⁺(aq)|Cu(s) ⇒ +0.34 V

Ag⁺(aq)|Ag(s) ⇒ +0.80 V

The most negative is Zn²⁺(aq)|Zn(s) ⇒ -0.76 V

From this, the most reducing agent is Zn. Zn(s) is oxidized to Zn²⁺ ions with the highest oxidation potential (0.76 V).

The given question is incomplete. The complete question is:

Compound X has a molar mass of 153.05 g/mol and the following composition:

element mass %

carbon 47.09%

hydrogen 6.59%

chlorine 46.33%

Write the molecular formula of X.

Answer: The molecular formula of X is [tex]C_6H_{10}Cl_2[/tex]

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 47.09 g

Mass of H = 6.59 g

Mass of Cl = 46.33 g

Step 1 : convert given masses into moles.

Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{47.09g}{12g/mole}=3.92moles[/tex]

Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.59g}{1g/mole}=6.59moles[/tex]

Moles of Cl =[tex]\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{46.33g}{35.5g/mole}=1.30moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{3.92}{1.30}=3[/tex]

For H = [tex]\frac{6.59}{1.30}=5[/tex]

For Cl =[tex]\frac{1.30}{1.30}=1[/tex]

The ratio of C : H: Cl= 3: 5 :1

Hence the empirical formula is [tex]C_3H_5Cl[/tex]

The empirical weight of [tex]C_3H_5Cl[/tex] = 3(12)+5(1)+1(35.5)= 76.5g.

The molecular weight = 153.05 g/mole

Now we have to calculate the molecular formula.

[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{153.05}{76.5}=2[/tex]

The molecular formula will be=[tex]2\times C_3H_5Cl=C_6H_{10}Cl_2[/tex]

Answer:

1.6x10¹¹ = Kc

Explanation:

For the reaction:

AgCl(s) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) + Cl⁻(aq)

Kc is defined as:

Kc = [Ag(CN)₂⁻] [Cl⁻] / [CN⁻]²

Ksp of AgCl is:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

Where Ksp is:

Ksp = [Ag⁺] [Cl⁻] = 1.6x10⁻¹⁰

In the same way, Kf of Ag(CN)₂⁻ is:

Ag⁺(aq) + 2CN⁻ ⇄ Ag(CN)₂⁻

Kf = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = 1.0x10²¹

The multiplication of Kf with Ksp gives:

[Ag⁺] [Cl⁻] *  [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = Ksp*Kf

[Ag(CN)₂⁻] [Cl⁻] / [CN⁻]² = Ksp*Kf

Obtaining the same expression of the first reaction

That means Ksp*Kf = Kc

1.6x10⁻¹⁰*1.0x10²¹ = Kc

Answer:

a. Rate = k×[A]

b. k = 0.213s⁻¹

Explanation:

a. When you are studying the kinetics of a reaction such as:

A + B → Products.

General rate law must be like:

Rate = k×[A]ᵃ[B]ᵇ

You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.

If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1

Rate = k×[A]¹[B]ᵇ

In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]

Rate = k×[A][B]⁰

b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:

Rate = k×[A]

0.320M/s = k×[1.50M]

Answer:

When balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.

Explanation:

Oxidation-reduction reactions or redox reactions are those in which an electron transfer occurs between the reagents. An electron transfer implies that there is a change in the number of oxidation between the reagents and the products.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The oxidation and reduction half-reactions, in a basic medium, adjust the oxygens and hydrogens as follows:

In the member of the half-reaction that presents excess oxygen, you add as many water molecules as there are too many oxygen. Then, in the opposite member, the necessary hydroxyl ions are added to fully adjust the half-reaction. Normally, twice as many hydroxyl ions, OH-, are required as water molecules have previously been added.

In short, you first adjust the oxygens with OH-, then you adjust the H with H₂O, and finally you adjust the charge with e-

So, when balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.

Answer:

c. balance the reaction as though under acidic conditions

Explanation:

When balancing redox reactions under basic conditions, a good technique is to first balance the reaction as though under acidic conditions. We then adjust the result to reflect the basic conditions.

Explanation:

According to your question,

no. a. ans would be like; chemical formula is defined as the an expression which determines no. and type of molecule of a compound.

b. no. ans; it tells that what type of compound is formed with the type and no. of atoms present in the atom.

c. no ans; the formulation of h2so4 states that it is acid named as hydrochloric acid which is formed by reacting of hydrogen (2 atoms ) ,sulpher (*1atom) and oxygen(4atoms).

Hope it helps...

Answer:

A precipitate will form, BaCO₃

Explanation:

When Ba²⁺ and CO₃²⁻ ions are in an aqueous media, BaCO₃(s), a precipitate, is produced following its Ksp expression:

Ksp = 5.1x10⁻⁹ = [Ba²⁺] [CO₃²⁻]

Where the concentrations of the ions are the concentrations in equilibrium

For actual concentrations of a solution, you can define Q, reaction quotient, as:

Q = [Ba²⁺] [CO₃²⁻]

If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form.

Actual concentrations of Ba²⁺ and CO₃²⁻ are:

[Ba²⁺] = [Ba(NO₃)₂] = 1.1x10⁻³ × (20.0mL / 100.0mL) = 2.2x10⁻⁴M

[CO₃²⁻] = [Na₂CO₃] = 8.4x10⁻⁴ × (80.0mL / 100.0mL) = 6.72x10⁻⁴M

100.0mL is the volume of the mixture of the solutions

Replacing in Q expression:

Q = [Ba²⁺] [CO₃²⁻]

Q = [2.2x10⁻⁴M] [6.72x10⁻⁴M]

Q = 1.5x10⁻⁷

As Q > Ksp

Answer:

ΔG°  = 1022. 8 kJ

Explanation:

ΔH° = –199 kJ/mol

ΔS° = –4.1 J/K·mol

T = 25°C = 25 + 273 = 298K (Converting to kelvin temperature)

ΔG° = ?

The relationship between these varriables are;

ΔG° = ΔH°  - TΔS°

ΔG° = –199 - 298 (–4.1)

ΔG° = -199 + 1221.8

ΔG°  = 1022. 8 kJ

To calculate ΔG, the following equation should be used:

ΔG° = ΔH°  - TΔS°

Given:

ΔH° = –199 kJ/mol,  

ΔS° = –4.1 J/K·mol

T=25+273K=298K  

Substitute the respective values:

ΔG° = ΔH°  - TΔS°

       =–199 kJ/mol-(298K×(-4.1 J/K·mol*(1KJ/1000J)))

       =-197.78kJ/mol

Thus, we can conclude the value of ΔG°=-197.78kJ/mol  

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answer is oxygen .

oxygen is an exception in octet rule

Among the following given elements,oxygen is an element which cannot have an expanded octet.

Expanded octet is a condition where an octet has more than 8 electrons and which is called as hyper-valency state. This concept is related to hybrid orbital theory and Lewis theory. Hyper-valent compounds  are not less common  and are of equal stability as the compounds which obey octet rule.

Expansion of octet is possible for elements from third period on wards only as they have low-lying empty d - orbitals which can accommodate more than eight electrons.

Expanded octet is not applicable to oxygen as it is second period of periodic table and has less than ten electrons and even does  not have the 2d -orbitals   due to which it  does not fulfill the criteria of an element to have an expanded octet.

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A 1.0-L buffer solution contains 0.100 mol  HC2H3O2 and 0.100 mol  NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

Answer:

The pH of this solution = 5.06  

Explanation:

Given that:

number of moles of CH3COOH = 0.100 mol

volume of the buffer solution = 1.0 L

number of moles of NaC2H3O2 = 0.100 mol

The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

we know that concentration in mole = Molarity/volume

Then concentration of [CH3COOH] = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex]  = 0.10 M

The chemical equation for this reaction is :

[tex]\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}[/tex]

The conjugate base is CH3COO⁻

The concentration of the conjugate base [CH3COO⁻] is  = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex]  

= 0.10 M

where the pka (acid dissociation constant)for CH3COOH = 4.74

If 0.035 mol of NaOH is added  to the original buffer, the concentration of NaOH added will be = [tex]\mathtt{ \dfrac{0.035 \ mol}{ 1.0 \ L }}[/tex] = 0.035 M

The ICE Table for the above reaction can be constructed as follows:

                  [tex]\mathtt{CH_3COOH \ \ \ + \ \ \ \ OH^- \ \ \to \ \ CH_3COO^- \ \ \ + \ \ \ H_2O}[/tex]

Initial             0.10               0.035         0.10                  -

Change        -0.035          -0.035       + 0.035              -

Equilibrium    0.065              0              0.135               -

By using  Henderson-Hasselbalch equation:

The pH of this solution = pKa + log [tex]\mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}[/tex]

The pH of this solution = 4.74 + log [tex]\mathtt{\dfrac{0.135}{0.065}}[/tex]

The pH of this solution = 4.74 + log (2.076923077 )

The pH of this solution = 4.74 + 0.3174

The pH of this solution = 5.0574

The pH of this solution = 5.06    to two decimal places