Answer:
41.9(w/w) %
Explanation:
Based on the reaction:
Na₂C₂O₄(s) + 2HCl(aq) → H₂C₂O₄(aq) + 2NaCl(aq)
Where 1 mole of sodium oxalate reacts with 2 moles of HCl
Moles of HCl solution to reach end point are:
44.15mL = 0.04415L ₓ (0.250mol / L) = 0.01104 moles of HCl
As 2 moles of HCl reacts per mole of Na₂C₂O₄:
0.01104mol HCl ₓ (1 mol Na₂C₂O₄ / 2 mol HCl) = 5.519x10⁻³ moles Na₂C₂O₄ are in the sample.
Molar mass of Na₂C₂O₄ is 134g/mol; thus, mass of 5.519x10⁻³ moles Na₂C₂O₄ is:
5.519x10⁻³ moles Na₂C₂O₄ ₓ (134g / mol) = 0.740g of Na₂C₂O₄ in the sample.
Thus, percent by mass of sodium oxalate in the sample is:
0.740g of Na₂C₂O₄ / 1.766g ₓ 100 =
41.9(w/w) %Balance the following equations: (c) H2(g)+I2(s)⟶HI(s)H2(g)+I2(s)⟶HI(s)
Answer: [tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Thus in the reactants, there are 2 atoms of hydrogen and 2 atoms of iodine .Thus there has to be 2 atoms of hydrogen and 2 atoms of iodine in the product as well. Thus a coefficient of 2 is placed in front of HI.
The balanced chemical reaction is:
[tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]
a ______ consumer is a heterotroph that directly eats an autotroph. A primary B. Quaterany C tertiary D secondary
A chemical reaction has the equation 2AgNO3 (aq) + Zn (s) 2Ag (s) + Zn(NO3)2 (aq). What type of reaction occurs between AgNO3 and Zn?
Answer: single replacement reaction
Explanation:
A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution. Thus one element should be different from another element.
A general single displacement reaction can be represented as :
[tex]X+YZ\rightarrow XZ+Y[/tex]
The reaction [tex]2AgNO_3(aq)+Zn(s)\rightarrow 2Ag(s)+Zn(NO_3)_2(aq)[/tex]
When zinc metal is added to aqueous silver nitrate, zinc being more reactive than silver displaces silver atom from its salt solution and lead to formation of zinc nitrate and silver metal.
A pentavalent cation atom has 20 and 15 neutrons as protons. Find the electron quantity and mass number respectively. (40 pts.) a) 20 and 15 b) 15 and 20 c) 15 and 35 d) 35 and 15 e) 10 and 20
Answer:
C.
Explanation:
Since the mass number is the number of protons and neutrons added together, the answer is 35. Since the questions are respectively electron quantity and mass number, the only answer choice with 35 as the second choice is C, so that is the correct answer.
How many moles of ammonia are in 0.175 L of a 6.50 M aqueous ammonia solution?
Answer:
Concentration (C) = number of moles (n) / volume (v)
Therefore
number of moles (n) = concentration × volume
Concentration = 6.50M
Volume =0.175 L = 0.175dm³
n = 6.50 × 0.175
n = 1.138 moles
Moles of ammonia in the solution is
1.138 moles.
Hope this helps
"Calculate the pH during the titration of 20.00 mL of 0.1000 M HF(aq) with 0.2000 M NaOH(aq) after 9.4 mL of the base have been added. Ka of hydrofluoric acid
Answer:
The answer is " 10.39"
Explanation:
Calculating acid moles:
[tex]= 0.02000 \ L \times 0.1000 \ M \\\\= 0.002000[/tex]
Calculating NaOH moles:
[tex]= 0.02012 \ L \times 0.1000 \ M \\\\= 0.002012[/tex]
calculating excess in OH- Moles:
[tex]= 0.002012 - 0.002000\\\\=0.000012[/tex]
calculating total volume:
[tex]= 20.00 + 20.12\\\\ = 40.12 mL \\\\= 0.04012 L[/tex]
[tex][OH-]= \frac{0.000012} { 0.0472}[/tex]
[tex]=0.00025 M[/tex]
[tex]pOH = - \log 0.00025[/tex]
= 3.6
[tex]pH = 14 - pOH[/tex]
= 10.39
Identify the elements that correspond to the following generalized electron configuration: (noble gas]ns2(n − 2) f6
Express your answer as the element symbol. If there is more than one answer, separate them by a comma.
Answer:
Samarium:
Electron configuration:
Samarium
Explanation:
Samarium is a chemical element that belongs to the lanthanoid series. The lanthanoids are the chemical elements that follow lanthanum. They are all known to possess 4f orbitals. The 4f electrons are found in the antepenultimate shell of the elements of the lanthanoid series and they do not take part in chemical bonding. They are neither removed in bonding nor do they take part in crystal field stabilization of lanthanoid complexes.
The electronic configuration of samarium is; 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f6 while the condensed, short hand electronic configuration is [Xe] 4f6 6s2. This corresponds to (noble gas]ns2(n − 2) f6 as required by the question, hence the answer provided above.
tank contains helium gas at 490 mm Hg, nitrogen gas at 0.75 atm and neon at 520 torr. What is the total pressure in atm? 2.1 atm 0.55 atm 1.5 atm 5.1 atm 51 atm
Answer:
2.1 atm
Explanation:
Before we get the total pressure, we have to ensure all the gases have the same pressure unit.
Nitrogen gas = 0.75 atm
Helium = 490mmHg
To convert mmHg to atm;
760 mmHg = 1 atm
490 = x
x = 460 / 760 = 0.645 atm
Neon = 520 torr
Converting torr to atm;
760 torr = 1 atm
520 torr = x
x = 520 / 760 = 0.684 atm
The total pressure is then given as;
0.75 + 0.684 + 0.645 = 2.1 atm
An unknown element, X, reacts with oxygen to form two oxides with the general formula X_2O_3 and X_2O_5. The pure element is strongly paramagnetic, and the oxides are amphoteric. Which of the following elements could be represented by X?a) Scb) Bc) Nd) Sbe) S
Answer:
The unknown element is Sb
Explanation:
The first thing we must note is that the unknown element must be a member of group 15 in the periodic table. This is clear from the fact that the two oxides formed are X2O3 and X2O5. This implies that the unknown element X must have a valency of 3 or 5. This corresponds to our knowledge that the outermost electron configuration of group 15 elements is ns2np3. Hence, group fifteen elements can have a valency of 3 or 5.
The electronic configuration of antimony is; [Kr]4d10 5s2 5p3. This implies that the atom is paramagnetic since there are three unpaired 5p electrons. The oxides of antimony are known to be amphoteric. An ampohoteric oxide reacts with both acid and base, hence the answer.
What do we call temperature changes caused by changes in air pressure?
Answer:
Fronts
Explanation:
For example, there are hot and cold fronts which cause the air to become warmer or cooler in a specific region!
Hope this helps! Please mark as brainiest!
Which of the following would have a fixed shape and volume? (3 points) nitrogen gas solid wood liquid water neon gas
Answer:
Solid Wood
Explanation:
Wood is like a solid block, whereas gases flow freely and liquids spread to fill the shape of their container.
Please let me know if I misunderstood the question, by the way.
If the pKaof HCHO2is 3.74 and the pH of an HCHO2/NaCHO2solution is 3.11, which of the following is true?
A. [HCHO2] < [NaCHO2]
B. [HCHO2] = [NaCHO2]
C. [HCHO2] << [NaCHO2]
D. [HCHO2] > [NaCHO2]
E. It is not possible to make a buffer of this pH from HCHO2 and NaCHO2
Answer:
D. [HCHO₂] > [NaCHO₂]
Explanation:
Formic acid, HCHO₂, is a weak acid that, in presence of its conjugate base, NaCHO₂ (CHO₂⁻), produce a buffer following H-H equation:
pH = pKa + log [CHO₂⁻] / [HCHO₂]
As pKa of the acid is 3.74 and pH of the solution is 3.11:
3.11 = 3.74 + log [CHO₂⁻] / [HCHO₂]
-0.63 = log [CHO₂⁻] / [HCHO₂]
0.2344 = [CHO₂⁻] / [HCHO₂]
A ratio [CHO₂⁻] / [HCHO₂] < 1, means:
[HCHO₂] > [CHO₂⁻]What is the name of Mn(CO3)2
Answer:
Mn is manganese and CO₃ is carbonate. Since the charge for CO₃ is -2 and the subscript is 2, the charge of Mn must be +4 so the answer is manganese (IV) carbonate.
Manganese (IV) carbonate is the name of Mn(CO[tex]_3[/tex])[tex]_2[/tex]. The only names used to identify salts are those of the cation or the anion.
The chemical formula of the anion (such as chloride or acetate) comes first in the name of a salt, which is followed with the identity of the cation (such as sodium or ammonium). They are created when acids and bases react, and they are always composed of either metal cations or cations made of ammonium. Manganese is Mn, and carbonate is CO[tex]_3[/tex]. The solution equals manganese (IV) carbonate since the charge for CO[tex]_3[/tex] is -2 but the subscript is 2, meaning that the charge of Mn has to be +4.
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The heat capacity of liquid water is 4.18 J/g·°C and the heat of vaporization is 40.7 kJ/mol. How many kilojoules of heat must be provided to convert 1.00 g of liquid water at 67°C into 1.00 g of steam at 100°C?
Answer:
The correct answer would be - 2.4KJ or, 2400J
Explanation:
Given:
heat capacity of liquid water - 4.18 J/g·°C
heat of vaporization - 40.7 kJ/mol
Mass of water = 1g
Moles of water = mass/molar mass
= 1g/18.016g
= 0.055 moles
Then,
Total heat required = q1(to raise the temperature to 100) + q2(change from the liquid phase to gas/steam)
= m *s*dt + moles * heat of vaporization
= (1g * 4.18 j/gc * (100-67)) + 0.055* 40.7 KJ
= 137.94J + 2.26KJ
=0.138KJ + 2.26KJ
=2.4KJ or, 2400J
Thus, the correct answer would be - 2.4KJ or, 2400J
Question 11: How does the energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital compare to the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital?
Answer:
Explanation:
The energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital is exactly same as the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital
Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C 6H 5COOH) and 0.20 M sodium benzoate (C 6H 5COONa), what will the pH of the solution be after the addition of 25.0 mL of 0.100M HCl? (K a (C 6H 5COOH) = 6.5 x 10 -5)
Answer:
pH = 4.05
Explanation:
The pH of the benzoic buffer can be determined using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where pKa is -logKa = 4.187
pH = 4.187 + log [Sodium Benzoate] / [Benzoic Acid]
Where [] can be understood as moles of each specie.
Thus, to find pH of the buffer we need to calculate moles of benzoic acid and sodium benzoate.
Initial moles:
Initial moles of benzoic acid and sodium benzoate are:
Acid: 250mL = 0.250L ₓ (0.250 moles / L) = 0.0625 moles of benzoic acid
Benzoate : 250mL = 0.20L ₓ (0.250 moles / L) = 0.050 moles of sodium benzoate
Moles after reaction:
Now, 0.0250L×(0.100mol/L) = 0.0025 moles of HCl are added to the buffer reacting with sodium benzoate, C₆H₅COONa, producing more benzoic acid, as follows:
HCl + C₆H₅COONa → C₆H₅COOH + NaCl
That means after reaction moles of both species are:
Benzoic acid: 0.0625 mol + 0.0025mol (Moles produced) = 0.065 moles
Sodium Benzoate: 0.050mol - 0.0025mol (Moles that react) = 0.0475 moles
Replacing in H-H equation:
pH = 4.187 + log [0.0475] / [0.065]
pH = 4.05
Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ∆G° = 349 kJ/mol
The given question is incomplete, the complete question is:
Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ΔG° = 349 kJ/mol
A) –309 kJ/mol
B) –329 kJ/mol
C) None of the above
D) –349 kJ/mol
E) –369 kJ/mol
Answer:
The correct answer is option D, that is, -349 kJ/mol.
Explanation:
Based on the given information, the reaction is:
NaBr (s) ⇔ Na (s) + 1/2 Br₂ (l), the ΔG° of the reaction given is 349 kJ per mole. In the given question, it is clearly mentioned that there is a need to determine the free energy of the formation of NaBr. Thus, there is a need to keep Na (s) and Br₂ (l) at the reactant side and NaBr (s) at the product side.
Therefore, there is a need to reverse the reaction and change the sign on ΔG.
Now the reaction will become,
Na (s) + 1/2 Br₂ (l) ⇔ NaBr (s), and the ΔG° will now become -349 kJ per mole. Hence, -349 kJ per mole is the free energy of the formation of NaBr (s).
The reaction: A + 3 B → D + F was studied and the following mechanism was determined. A + B C (fast) C + B → D + E (slow) E + B → F (very fast) The species, C, is properly described as
Answer:
Intermediate.
Explanation:
Hello,
In this case, we can rewrite the steps as:
[tex]A + B \rightarrow C\ \ (fast)\\\\C + B \rightarrow D + E\ \ (slow)\\\\E + B \rightarrow F \ \ (very fast)[/tex]
Thus, we can notice that in the fast step, C is present as a product but after that is consumed in the slow step, for that reason, and by cause of its formation-consumption behavior, it is properly described as an intermediate as it is not neither a starting-up substance (reactant in the first step) nor a final substance (product in the final step).
Best regards.
An aqueous solution of nitric acid is standardized by titration with a 0.110 M solution of calcium hydroxide. If 21.1 mL of base are required to neutralize 23.8 mL of the acid, what is the molarity of the nitric acid solution? ? M Nitric acid
Answer:
[tex]\large \boxed{\text{0.195 mol/L}}[/tex]
Explanation:
(a) Balanced equation
2HNO₃ + Ca(OH)₂ ⟶ Ca(NO₃)₂ + 2H₂O
(b) Moles of Ca(OH)₂
[tex]\text{Moles of Ca(OH)}_{2} = \text{21.1 mL Ca(OH)}_{2} \times \dfrac{\text{0.110 mmol Ca(OH)}_{2}}{\text{1 mL Ca(OH)}_{2}}\\= \text{2.321 mmol Ca(OH)}_{2}[/tex]
(c) Moles of HNO₃
The molar ratio is 2 mol HNO₃:1 mol Ca(OH)₂
[tex]\text{Moles of HNO}_{3} = \text{2.321 mmol Ca(OH)}_{2} \times\dfrac{\text{2 mmol HNO}_{3}}{\text{1 mmol Ca(OH)}_{2}}= \text{4.642 mmol HNO}_{3}[/tex]
(d) Molar concentration of HNO₃
[tex]c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V}\\\\c= \dfrac{\text{4.642 mmol}}{\text{23.8 mL}} = \text{0.195 mol$\cdot$L$^{-1}$}\\\\\text{The molar concentration of the Ca(OH)$_{2}$ is $\large \boxed{\textbf{0.195 mol/L}}$}[/tex]
Add distilled water to the beaker until the volume
totals 15 mL.
Record the amount of oil that dissolved.
Answer:
i guess oil never dissolve in water. As like dissolve like. water is polar so it dissolves only polar substances
Explanation:
Answer:
None
Explanation:
Answer on Edge 2022
Can a catalyst change an exothermic reaction into an endothermic reaction or vice versa? Please explain your answer.
Answer:
A catalyst cannot change an exothermic reaction into an endothermic reaction or vice versa.
Explanation:
Catalyst is basically a substance that enables a chemical reaction to occur at a faster rate as compared to the reaction without catalysis. It lowers the activation energy and temperature for a chemical reaction and a catalyst itself does not goes through any permanent chemical change. This means it does not get used in the process.
Exothermic and endothermic are the chemical reaction. Exothermic reactions absorb energy. This energy is absorbed in the form of heat. When the energy is released in the form of heat then this reaction is called endothermic. So one absorbs the heat and the other releases it.
As we know that the catalyst does not undergo change at the end of the reaction so the energy or heat whether is absorbed or emitted or you can say whether the reaction is exothermic or endothermic, the total energy stays unchanged during the reaction. So with and without a catalyst, if both have same reactants and products and the difference in enthalpy between products and reactants will be the same.
Which of the following best describes a salt bridge? a) A pathway composed of salt water that ions pass through. b) A pathway between the cathode and anode in which ions are reduced. c) A pathway by which counterions can flow between the half-cells with the solutions in the half-cell completely mixing.
Answer: A
Explanation:
Which statement best describes covalent bonding?
Answer:
Option C. Electrons are shared between two atoms
Explanation:
Covalent bonding is a type of bonding which exist between two non metals.
In this bonding, electrons are shared between the two atoms involved in order to attain a stable octet configuration.
This can be seen when hydrogen atom combine with chlorine atom to form hydrogen chloride as shown below:
H + Cl —> HCl
Hydrogen has 1 electron in it's outmost shell and it requires 1 electron to attain a stable configuration.
Chlorine has 7 electrons in it's outmost shell and requires 1 electron to attain a stable configuration.
During bonding, both hydrogen and chlorine will contribute 1 electron each to form bond, thereby attaining a stable configuration. The bond formed in this case is called covalent bond as both atoms involved shared electron to attain a stable configuration.
C. Electrons are shared between two atoms.
What is Covalent Bonding?
A covalent bond is fashioned among non-metals which have comparable electronegativities. Neither atom is "strong" sufficient to draw electrons from the other.It is formed when pairs of electrons are shared by atoms.Atoms will covalently bond with different atoms that allows you to benefit extra stability, that is gained through forming a complete electron shell. By sharing their outer most (valence) electrons, atoms can replenish their outer electron shell and gain stability.For example:
In H₂ molecule; there is a covalent bond formation between two hydrogen atoms as the electron from each hydrogen atom is shared leading to the formation of hydrogen molecule.
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A 5.22 × 10−3−mol sample of HY is dissolved in enough H2O to form 0.088 L of solution. If the pH of the solution is 2.37, what is the Ka of HY?
Answer:
3.07 × 10⁻⁴
Explanation:
Step 1: Calculate the concentration of H⁺
We will use the definition of pH.
[tex]pH = -log [H^{+} ]\\\[ [H^{+} ] = antilog -pH = antilog -2.37 = 4.27 \times 10^{-3} M[/tex]
Step 2: Calculate the concentration of HY
5.22 × 10⁻³ mol of HY are dissolved in 0.088 L. The concentration of the acid (Ca) is:
[tex]Ca = \frac{5.22 \times 10^{-3} mol }{0.088L} = 0.0593M[/tex]
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
[tex]Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(4.27 \times 10^{-3} )^{2} }{0.0593} = 3.07 \times 10^{-4}[/tex]
Dissolving NaOH(s) in water is exothermic. Two calorimetry experiments are set up. Experiment 1: 2 g of NaOH are dissolved in 100 mL of water Experiment 2: 4 g of NaOH are dissolved in 200 mL of water Which of the following statements is true?a. both temperature changes will be the sameb. the second temeprature change will be approximately twice the firstc. the second temperature change will be approximately four times the firstd. the second temperature change will be approximately one-half of the firste. the second temperature change will be approximately one-fourth the first
Answer:
a. both temperature changes will be the same
Explanation:
When sodium hydroxide (NaOH) is dissolved in water, a determined amount is released to the solution following the equation:
Q = m×C×ΔT
Where Q is the heat released, m is the mass of the solution, C is the specific heat and ΔH is change in temperature.
Specific heat of both solutions is the same (Because the solutions are in fact the same). Specific heat = C.
m is mass of solutions: 102g for experiment 1 and 204g for experiment 2.
And Q is the heat released: If 2g release X heat, 4g release 2X.
Thus, ΔT in the experiments is:
Experiment 1:
X / 102C = ΔT
Experiment 2:
2X / 204C = ΔT
X / 102C = ΔT
That means,
a. both temperature changes will be the same
Devise a detailed experimental procedure to purify ~ 20 grams of benzoic acid that is contaminated with sodium chloride. Justification of the steps (including solubility calculations) that are included in the procedure. In other words, explain why the steps are being included.
Answer:
Based on the difference in solubility one can perform the process of purification of the benzoic acid contaminated with sodium chloride. The benzoic acid does not get soluble in cold water, while the sodium chloride is soluble in cold water.
Thus, for separation, the supplementation of cold water can be done into the mixture in the experiment of purifying benzoic acid from sodium chloride. In the process, the mixture is placed on the ice bath and is stirred well, in the end, the solution is filtered. The filtrate contains sodium chloride and on the filter paper pure benzoic acid is collected.
Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the following endothermic reaction. 2N2(g) + O2(g)2N2O(g) Clear All > 0 Hrxn < 0 Srxn = 0 Grxn > 0 low T, < 0 high T Suniverse < 0 low T, > 0 high T
Answer:
∆H > 0
∆Srxn <0
∆G >0
∆Suniverse <0
Explanation:
We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.
Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.
The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.
Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.
If 2 moles of helium undergo a temperature increase of 100 K at constant pressure, how much energy has been transferred to the helium as heat
Answer:
[tex]Q=4154J[/tex]
Explanation:
Hello,
In this case, the involved heat in this heating process is considered to be computed via:
[tex]Q=nCp\Delta T[/tex]
Whereas we assume a constant molar specific heat of helium which is 20.77 J/(mol*K), thus, the transferred energy in the form of heat turns out:
[tex]Q=2mol*20.77\frac{J}{mol*K} *100K\\\\Q=4154J[/tex]
Regards.
Calculate the pH for the following 1.0M weak acid solutions:a. HCOOH Ka = 1.8 x 10-4 [
Answer: pH=2.38
Explanation:
To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.
HCOOH ⇄ H⁺ + HCOO⁻
I 1.0M 0 0
C -x +x +x
E 1.0-x x x
For the steps below, refer to the ICE chart above.
1. Since we were given the initial of HCOOH, we can fill this into the chart.
2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.
3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.
4. We were given the Kₐ of the solution. We know [tex]K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex].
5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.
[tex]1.8*10^-^4 =\frac{x^2}{0.1-x}[/tex]
6. Once we plug this into the quadratic equation, we get x=0.00415.
7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).
-log(0.00415)=2.38
Our pH for the weak acid solution is 2.38.
Calculate the pH of mixing 24 mL of 1M acetic acid with 76 mL of 1M sodium acetate. For the purpose of this calculation, assume the Ka of acetic acid is 1.8 X 10-5. You must include units to obtain full credit. You must show all your work to obtain any credit.
Answer:
pH = 5.24
Explanation:
Mixture of acetic acid with acetate ion is a buffer (Mixture of a weak acid with its conjugate base). The pH of a buffer can be determined using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
Where pKa is -log Ka = 4.74; [A⁻] is the concentration of conjugate base (Acetate ion) and [HA] is molar concentration of the weak acid.
Concentration of the acetic acid in the 100mL≡0.1L (76mL + 24mL) solution is:
[HA] = 0.024L ₓ (1mol / L) / 0.1L = 0.24M
[A⁻] = 0.076L ₓ (1mol / L) / 0.1L = 0.76M
Replacing in H-H equation:
pH = 4.74 + log₁₀ [0.76M] / [0.24M]
pH = 5.24