I've been stuck on a homework problem for a while now. Can someone help me get to the answer?
An acid has a Ka of 2.9 x 10^-6. For 100 mL of a 0.2 M acid solution being titrated with a 0.05 M NaOH solution, determine the pH after the following volumes of the NaOH solution are added. SHOW ALL WORK.
A. 9mL
B. 100 mL
C. 400 mL
D. 500 mL
Don't use ChatGPT, I can tell if you do, please.
Answers
A) pH after adding 9 mL of NaOH is 6.91 ; B) pH after adding 100 mL of NaOH is 13.18 ; C) pH after adding 400 mL of NaOH is 8.42 ; D) pH after adding 500 mL of NaOH is 12.
What is pH?pH is used to find out the the basicity and acidity of a solution
HA + NaOH -> NaA + H₂O
Ka = [NaA][H₃O⁺] / [HA]
moles of acid = initial concentration x initial volume = 0.2 M x 0.1 L = 0.02 moles
A.) We add 9 mL of 0.05 M NaOH to acid solution. The total volume of the solution is now 100 mL + 9 mL = 0.109 L.
moles of NaOH added = concentration x volume = 0.05 M x 0.009 L = 4.5 x 10⁻⁴ moles
moles of acid remaining = initial moles - moles of NaOH added = 0.02 moles - 4.5 x 10⁻⁴ moles = 0.01955 moles
moles of NaA formed = moles of NaOH added = 4.5 x 10⁻⁴ moles
moles of H₃O⁺ produced = moles of NaOH added = 4.5 x 10⁻⁴ moles
moles of HA remaining = moles of acid remaining - moles of NaA formed = 0.01905 moles
Ka = [NaA][H₃O] / [HA]
[ H₃O⁺] = Ka x [HA] / [NaA]
[ H₃O⁺] = 2.9 x 10⁻⁶ x 0.01905 / 4.5 x 10⁻⁴ = 1.22 x 10⁻⁷ M
pH = -log[H₃O⁺] = -log(1.22 x 10⁻⁷) = 6.91
Therefore, pH after adding 9 mL of NaOH is 6.91.
B.) 100 mL of 0.05 M NaOH, so all the acid has been neutralized.
moles of NaOH added = concentration x volume = 0.05 M x 0.1 L = 0.005 moles
moles of NaOH remaining = moles of NaOH added - moles of acid = 0.005 moles - 0.02 moles = -0.015 moles (negative value indicates excess NaOH)
[OH-] = moles of excess NaOH / total volume = 0.015 moles / 0.1 L = 0.15 M
pOH = -log[OH-] = -log(0.15) = 0.82
So pH= 14 -pOH = 14 -0.82= 13.18
Therefore, pH after adding 100 mL of NaOH is 13.18.
C.) We have added a total of 400 mL of 0.05 M NaOH.
moles of NaOH added = concentration x volume = 0.05 M x 0.4 L = 0.02 moles
moles of NaOH remaining = moles of NaOH added - moles of acid = 0.02 moles - 0.02 moles = 0 moles
[NaA] = moles of NaA / total volume = 0.02 moles / 0.5 L = 0.04 M
Kb = Kw / Ka = 1.0 x 10⁻¹⁴ / 2.9 x 10⁻⁶ = 3.45 x 10⁻⁹
[OH-] = √(Kb x [NaA]) = √(3.45 x 10⁻⁹ x 0.04) = 2.61 x 10⁻⁶ M
pOH = -log[OH-] = -log(2.61 x 10⁻⁶) = 5.58
So, pH= 14 -pOH = 14 -5.58 = 8.42
Therefore, pH after adding 400 mL of NaOH is 8.42.
D.) We have added a total of 500 mL of 0.05 M NaOH.
moles of NaOH added = concentration x volume = 0.05 M x 0.5 L = 0.025 moles
moles of NaOH remaining = moles of NaOH added - moles of acid = 0.025 moles - 0.02 moles = 0.005 moles
[OH-] = moles of excess NaOH / total volume = 0.005 moles / 0.5 L = 0.01 M
pOH = -log[OH-] = -log(0.01) = 2
So pH =14 -pOH =14 -2=12
Therefore, pH after adding 500 mL of NaOH is 12.
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The classification is:
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What volume of 18.0 M H2SO4 is needed to contain 2.45 moles of H2SO4?
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Please answer the questions in the picture below
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Answer b
Explanation:
ain the energy transformation that occurs when a hair dryer is plugged in and turned on.
Group of answer choices
Chemical energy transforms into electrical and sound energy.
Electrical energy transforms into mechanical and thermal energy.
Mechanical energy transforms into chemical and electrical energy.
Thermal energy transforms into electrical and gravitational energy.
Answers
Answer:
Mechanical
Explanation:
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A cell membrane is more flexible than a brick wall. Why might many cells benefit from a flexible cell membrane?
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A cell membrane is more flexible than a brick wall. Why might many cells benefit from a flexible cell membrane because it allows them to survive in and out of the cell.
The cell is surrounded by a membrane called as cell membrane . cell membrane is a semipermeable membrane. semipermeable membrane is the type of membrane which allows solvent particles to pass through it . or some molecules or ion . Therefore , many cells benefited by cell membrane as they can pass through it.
Thus, A cell membrane is more flexible than a brick wall. Why might many cells benefit from a flexible cell membrane because it allows them to survive in and out of the cell.
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what happens to the temperatures when we have direct sunlight vs indirect sunlight?
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Explanation:
When we have direct sunlight, the temperature is higher compared to indirect sunlight.
Direct sunlight is usually overhead whereas indirect sunlight is at an angle to the surface.
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of solid silver chromate will precipitate
when 150. mL of 0.500 M silver nitrate
solution is added to excess potassium
chromate.
2AgNO3(aq)
How many moles of AgNO3 are present
in 150. mL of 0.500 M AgNO3?
+ K₂ CrO4 (aq) → Ag₂ CrO4(s) + 2KNO3(aq)
Answers
Approximately 12.45 grams of solid silver chromate will precipitate when 150 mL of 0.500 M silver nitrate solution is added to excess potassium chromate.
To determine the number of moles of AgNO3 present in 150 mL of a 0.500 M AgNO3 solution, we can use the formula:
moles = concentration × volume
Given:
Concentration of AgNO3 solution = 0.500 M
Volume of AgNO3 solution = 150 mL
First, we need to convert the volume from milliliters (mL) to liters (L) since the concentration is given in moles per liter (M).
1 L = 1000 mL
Therefore, the volume of the AgNO3 solution in liters is:
150 mL × (1 L / 1000 mL) = 0.150 L
Now we can calculate the moles of AgNO3 using the formula:
moles = concentration × volume
moles = 0.500 M × 0.150 L
moles = 0.075 mol
So, there are 0.075 moles of AgNO3 present in 150 mL of the 0.500 M AgNO3 solution.
Now, let's proceed to determine how many grams of solid silver chromate (Ag2CrO4) will precipitate when the AgNO3 solution reacts with excess potassium chromate (K2CrO4).
From the balanced chemical equation:
2AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2KNO3(aq)
We can see that the molar ratio between AgNO3 and Ag2CrO4 is 2:1. Therefore, for every 2 moles of AgNO3, we will form 1 mole of Ag2CrO4.
Since we have 0.075 moles of AgNO3, we can calculate the moles of Ag2CrO4 formed:
moles of Ag2CrO4 = 0.075 mol / 2 = 0.0375 mol
To determine the mass of Ag2CrO4, we need to multiply the moles by its molar mass. The molar mass of Ag2CrO4 is calculated by summing the atomic masses of each element in the compound:
Ag2CrO4 = 2(Ag) + 1(Cr) + 4(O) = 2(107.87 g/mol) + 1(52.00 g/mol) + 4(16.00 g/mol) = 331.87 g/mol
mass of Ag2CrO4 = moles of Ag2CrO4 × molar mass of Ag2CrO4
mass of Ag2CrO4 = 0.0375 mol × 331.87 g/mol = 12.45 g
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which reagent (reactant) is limiting when 1.85 mol of sodium hydroxide and 1.00 mol carbon dioxide are allowed to react? how many moles of sodium carbonate can be produced? how many moles of the excess reactant remain after the completion of the reaction?
Answers
In the given chemical reaction \(\rm NaOH + CO_2 \rightarrow Na_2CO_3 + H_2O\), the limiting reagent is carbon dioxide and \(\rm Na_2CO_3\) that can be produced is 1.00 mol, and 0.85 mol of NaOH remains unreacted after the completion of the reaction.
The limiting reagent is the reactant that gets completely consumed in a chemical reaction, thereby limiting the amount of product that can be formed.
The balanced chemical equation for the reaction between sodium hydroxide and carbon dioxide is:
\(\rm NaOH + CO_2 \rightarrow Na_2CO_3 + H_2O\)
To determine which reactant is limiting, we need to calculate the number of moles of sodium hydroxide and carbon dioxide present and compare the mole ratios of the reactants in the balanced equation.
Moles of NaOH = 1.85 mol
Moles of \(\rm CO_2\) = 1.00 mol
The balanced equation shows that the mole ratio of NaOH to \(\rm CO_2\) is 2:1, which means that 2 moles of NaOH react with 1 mole of \(\rm CO_2\) .
Using the mole ratio, we can calculate the maximum number of moles of Na2CO3 that can be produced by each reactant:
Moles of NaOH / 2 = 0.925 mol \(\rm Na_2CO_3\)
Moles of\(\rm CO_2 \times 1\) /1 = 1.00 mol \(\rm Na_2CO_3\)
Since the calculated number of moles of \(\rm Na_2CO_3\) from \(\rm CO_2\) is less than the calculated number of moles of \(\rm Na_2CO_3\) from NaOH, \(\rm CO_2\) is the limiting reactant.
According to the balanced equation, 1 mole of \(\rm CO_2\) reacts to form 1 mole of \(\rm Na_2CO_3\) . Therefore, the number of moles of \(\rm Na_2CO_3\) produced is 1.00 mol.
To determine the amount of excess reactant remaining after the reaction, we need to calculate the amount of NaOH that did not react.
Moles of NaOH remaining = Moles of NaOH - (Moles of \(\rm Na_2CO_3\) produced x 2)
= 0.85 mol NaOH
Therefore, 0.85 mol of NaOH remains unreacted after the completion of the reaction.
In conclusion, carbon dioxide is the limiting reactant, and 1.00 mol of \(\rm Na_2CO_3\) can be produced. 0.85 mol of NaOH remains unreacted after the completion of the reaction.
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The question is incomplete. The complete question is:
Sodium hydroxide reacts with carbon dioxide as follow:
\(\rm NaOH + CO_2 \rightarrow Na_2CO_3 + H_2O\)
Which reagent (reactant) is limiting when 1.85 mol of sodium hydroxide and 1.00 mol carbon dioxide are allowed to react? How many moles of sodium carbonate can be produced? How many moles of the excess reactant remain after the completion of the reaction?
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Answers
Answer:
A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)
Explanation:
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The consituent elements of calcium carbonate, CaCO₃, in their standard states (States you will find this pure elements in nature), are:
Ca(s), C(s) and O₂(g)
That means, the equation that represents standard enthalpy of CaCO₃ is:
A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)Is the equation that has ΔH° = -1207kJ/mol
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The solver feature is a mathematical tool that is designed to find the optimal solution to a problem within certain constraints. The goal of the solver feature is to maximize or minimize the objective function while satisfying the constraints set by the user. The solver can handle linear and nonlinear problems, and can be used in a variety of applications, including engineering, finance, and operations management.
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Answer:
a cloud of adult stars that floats through space where protostars are born.
Explanation:
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As < Ge < Sn < In
As < Ge < In < Sn
Ge < As < In < Sn
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Answers
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Q-3 Determine the fugacity in atm for pure ethane at 310 K and 20.4 atm and change in the chemical potential between this state and a second state od ethane where temperature is constant but pressure is 24 atm.
Answers
The fugacity in atm for pure ethane at 310 K and 20.4 atm is given by the equation: f = 20.4 exp (-Δg1/RT). The change in chemical potential between this state and a second state of ethane where the temperature is constant but the pressure is 24 atm is -0.0911RT.
Fugacity is a measure of the escaping tendency of a component in a mixture, which is defined as the pressure that the component would have if it obeyed ideal gas laws. It is used as a correction factor in the calculation of equilibrium constants and thermodynamic properties such as chemical potential. Here we need to determine the fugacity in atm for pure ethane at 310 K and 20.4 atm and the change in the chemical potential between this state and a second state of ethane where the temperature is constant but the pressure is 24 atm. So, using the formula of fugacity: f = P.exp(Δu/RT) Where P is the pressure of the system, R is the gas constant, T is the temperature of the system, Δu is the change in chemical potential of the system. Δu = RT ln (f / P)The chemical potential at the initial state can be calculated using the ideal gas equation as: PV = nRT
=> P
= nRT/V
=> 20.4 atm
= nRT/V
=> n/V
= 20.4/RT The chemical potential of the system at the initial state is:
Δu1 = RT ln (f/P)
= RT ln (f/20.4) Also, we know that for a pure substance,
Δu = Δg. So,
Δg1 = Δu1 The change in pressure is 24 atm – 20.4 atm
= 3.6 atm At the second state, the pressure is 24 atm.
Using the ideal gas equation, n/V = 24/RT The chemical potential of the system at the second state is: Δu2 = RT ln (f/24) = RT ln (f/24) The change in chemical potential is Δu2 – Δu1 The change in chemical potential is
Δu2 – Δu1 = RT ln (f/24) – RT ln (f/20.4)
= RT ln [(f/24)/(f/20.4)]
= RT ln (20.4/24)
= - 0.0911 RT Therefore, the fugacity in atm for pure ethane at 310 K and 20.4 atm is:
f = P.exp(Δu/RT)
=> f
= 20.4 exp (-Δu1/RT)
=> f
= 20.4 exp (-Δg1/RT) And, the change in the chemical potential between this state and a second state of ethane where the temperature is constant but pressure is 24 atm is -0.0911RT. Therefore, the fugacity in atm for pure ethane at 310 K and 20.4 atm is given by the equation: f = 20.4 exp (-Δg1/RT). The change in chemical potential between this state and a second state of ethane where the temperature is constant but the pressure is 24 atm is -0.0911RT.
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In your own words, what does the law of
conservation of energy tell us? Hint: It is
VERY similar to the law of conservation
of matter (we studied that in the first
semester).
Answers
Answer:
The law of conservation of mass States that matter can never be created or destroyed during c
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Answers
Answer:
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Explanation:
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A molecule's three-dimensional structure, with simple lines representing plane links, wedge-shaped lines representing bonds facing the observer, and dashed lines representing bonds Stereochemistry facing the viewer's opposite.
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Answers
• The balanced equation for the reaction is given by:
CH3CH2OH + O2 → CH3COOH+ H2O• From the above reaction we can see that:
1 mole of ethanol (CH3Ch2OH) produces 1 mol of acetic acid(CH3COOH)
so . x mole of ethanol will produce 0.9mol of acetic acid ....(cross multiply)
xmol ethanol * 1mol acetic = 1mol ethamol* 0.9molacetic
∴ xmol ethanol = 1*0.9 /1
= 0.90 mol
• This means that, 0.90 mol of ethanol, is needed to produce 0.9mol acetic acid,.
What unit is used for atomic mass
Answers
Answer is daltons
Atomic weight is measured in atomic mass units (amu), also called daltons.
Atomic Mass
Masses of individual atoms are very, very small. Using a modern device called a mass spectrometer, it is possible to measure such minuscule masses. An atom of oxygen-16, for example, has a mass of 2.66 × 10 -23 g. While comparisons of masses measured in grams would have some usefulness, it is far more practical to have a system that will allow us to more easily compare relative atomic masses. Scientists decided on using the carbon-12 nuclide as the reference standard by which all other masses would be compared. By definition, one atom of carbon-12 is assigned a mass of 12 atomic mass units (amu). An atomic mass unit is defined as a mass equal to one twelfth the mass of an atom of carbon-12. The mass of any isotope of any element is expressed in relation to the carbon-12 standard. For example, one atom of helium-4 has a mass of 4.0026 amu. An atom of sulfur-32 has a mass of 31.972 amu.
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A 5 Complete the sentences using words from the list. You may use them once, more than once, or not at all.
increases；decreases；stays the same
a) When liquid water boils, the distance between the particles ____
b) When liquid water boils, the strength of the attractive forces between the particles ___
c) When steam condenses, the speed of movement of the particles ___
Answers
Answer
Number 1. increase
number 2 . decrease
number 3. decrease
which are the only two elements liquid at room temperature?
Answers
Answer:
mercury and bromine are the only two elements
The solubility of sodium chloride decreases as temperature increases.
True
False
Answers
Yes the increase in temperature of sodium chloride will cause a decrease in it solubility.
Answer is True
Solubility of Sodium ChlorideSodium chloride is a type of salt whose solubility decreased with increased temperature, this is quite unusual to other solute.
With the case of Sodium Chloride, this principle is applied in purifications like re-crystallization which is used to recover solute in solvents
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