If a buffer solution is 0.120 M in a weak acid (Ka = 2.0 × 10-5) and 0.440 M in its conjugate base, what is the pH?
To find the pH of the buffer solution, the Henderson-Hasselbalch equation is used with the provided concentrations of weak acid and conjugate base, yielding a pH value of approximately 5.26.
Explanation:To calculate the pH of the buffer solution containing a weak acid and its conjugate base, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
Given that the weak acid concentration is 0.120 M its conjugate base concentration is 0.440 M, and the Ka for the weak acid is 2.0 × 10^-5, we first calculate pKa:
pKa = -log(Ka) = -log(2.0 × 10^-5) = 4.70
Next, we use the Henderson-Hasselbalch equation:
pH = 4.70 + log(0.440/0.120) = 4.70 + log(3.667) ≈ 4.70 + 0.564 = 5.264
The pH of the buffer solution is therefore approximately 5.26.
Final answer:
To calculate the pH of the buffer solution, first determine the pKa from the given Ka and then apply the Henderson-Hasselbalch equation using the concentrations of the weak acid and its conjugate base. The pH of the given buffer solution is approximately 5.27.
Explanation:
The pH of a buffer solution can be determined using the Henderson-Hasselbalch equation, which is pH = pKa + log(·[A−]/[HA]·), where [A−] is the concentration of the conjugate base and [HA] is the concentration of the acid. In this case, the weak acid has a given Ka value of 2.0 × 10⁻µ, which allows us to calculate its pKa as the negative logarithm of Ka (pKa = -log(Ka)).
First, calculate the pKa:
pKa = -log(Ka)
= -log(2.0 × 10⁻µ)
= 4.70
Now, apply the Henderson-Hasselbalch equation:
pH = pKa + log([A−]/[HA])
= 4.70 + log(0.440 M/0.120 M)
= 4.70 + log(3.67)
= 4.70 + 0.565
= 5.265
Therefore, the pH of the buffer solution is approximately 5.27.
Find the number of moles of water that can be formed if you have 182 mol of hydrogen gas and 86 mol of oxygen gas.
The most probable mechanism of reactivity for the solvolysis of 2-chloro-norbornane is
The rate determining step for the reactivity for the solvolysis of 2-chloro-norbornane depends only on the decomposition of a single molecular species which is the 2-chloro-norbornane itself. For unimolecular reactions, the mechanism pathway being followed is that of an SN1 mechanism.
Answer:
SN1 mechanism
Given the formula m1v1=m2v2, where m indicates concentration and v indicates volume, which equation represents the correct way to find the concentration of the dilute solution (m2)?
M₂ = (M₁V₁) / V₂
Further explanationDilution represents the addition of a solvent (water) without adding solutes. In dilution, the mole of the solute remains, so the concentration of the solution will drop.
When calculating dilution factors, the units of volume and concentration must remain consistent.
Dilution calculations can be typically performed following the formula:
[tex]\boxed{ \ M_1V_1 = M_2V_2 \ }[/tex].
with,
V₁ and V₂ as volume before and after dilutionM₁ and M₂ as the molarity of the solution before and after dilution.The equation which represents the correct way to find the concentration of the dilute solution (M₂) is
[tex]\boxed{ \ M_2 = \frac{M_1V_1}{V_2} \ }[/tex].
_ _ _ _ _ _ _ _ _ _
Example:
How much must be dissolved to carry out 5 liters of 0.4 M methanol solution? How much water do you require adding?
M₁ = 25 MV₂ = 5 LM₂ = 0.4 MV₁ = ?[tex]\boxed{ \ V_1 = \frac{M_2V_2}{M_1} \ }[/tex]
[tex]\boxed{ \ V_1 = \frac{0.4 \times 5}{25} \ }[/tex]
[tex]\boxed{ \ V_1 = \frac{0.4}{5} \ }[/tex]
[tex]\boxed{ \ V_1 = \frac{0.8}{10} \ }[/tex]
[tex]\boxed{ \ V_1 = \frac{8}{100} \ }[/tex]
Thus we have 0.08 L or 80 mL of 25 M methanol solution.The amount of water that needs to be added is 5 L - 0.08 L = 4.92 L or 492 mL._ _ _ _ _ _ _ _ _ _
Notes:
A solution consists precisely of a solute and a solvent. There are several ways to properly express the concentration of a solution, one of which is molarity. Molarity is also known as molar concentration with the symbol unit M or molar or mole/L.The molarity (M) of a solution is calculated by allowing the moles of solute and divided by the number of liters of solution.[tex]\boxed{ \ molarity = \frac{moles \ of \ solute}{liters \ of \ solution} \ }[/tex]
Learn moreTo make a 0.500 M solution, one could take 0.500 moles of solute and add? https://brainly.com/question/10278982 The molality and mole fraction of water https://brainly.com/question/10861444 How many liters of the 50% solution and how many liters of the 90% solution will be used? https://brainly.com/question/13034221M₂ = (M₁V₁) / V₂ is the equation which represents the correct way to find the concentration of the dilute solution.
In chemistry, concentration is the ratio of solute to solvent or solution in a given volume. The amount of material in various solutions is quantified and compared using this crucial characteristic. Chemistry requires concentration in many different areas, such as chemical reactions, analyte detection, and solution preparation. Concentration can be expressed in a number of different ways, including molarity (M), mass/volume percent (%), parts per million (ppm), and mole fraction. Depending on the needs of the experiment or application, each approach offers a distinctive viewpoint on the solute concentration in a solution and is applied in various situations.
M₂ V₂= (M₁V₁)
M₂ = (M₁V₁) / V₂
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NaCl + H2O → NaOH + Cl2 + H2 How many grams of chlorine gas, Cl2 are given off if 7.5 grams of sodium chloride, NaCl, are decomposed? Is this equation balanced?
Rank the size of a change in temperature of one degree Fahrenheit, one degree Celsius, and one kelvin. In other words, if a thermometer were to show that the temperature outside increased by these amounts, which change would feel the largest? If any of the options are the same magnitude, stack them above one another. Rank the sizes of one degree in each scale from largest to smallest. To rank items as equivalent, overlap them.
Draw the compound that would produce 4-ethyl-3-hexanol in the presence of a nickel catalyst and hydrogen.
The compound that produces 4-ethyl-3-hexanol in the presence of nickel and hydrogen is [tex]\boxed{{\text{4 - ethylhexan - 3 - one}}}[/tex]. (Refer to the attached image)
Further Explanation:
The reduction reaction is a process in organic reactions where hydrogen atoms with or without the presence of catalyst like (Ni, Pt, and Pd.) are added to organic molecules like alcohols, phenols, aldehyde, and alkenes. It is also defined as a reaction where a less electronegative species is added to a more electronegative species.
Reduction of aldehyde or ketone compounds in the presence of nickel catalysts and hydrogen gives alcohol as the product.
A general reduction reaction is as follows:
[tex]{\text{RCHO}}\;{\text{ + }}\;{{\text{H}}_{\text{2}}}{\text{/Ni}} \to {\text{RC}}{{\text{H}}_{\text{2}}}{\text{OH}}[/tex]
In the hydrogenation reactions of aldehyde, the reduction of the carbonyl group takes place. The reaction of an aldehyde with hydrogen gas in presence of nickel catalyst leads to the formation of a primary alcohol, whereas the reaction of ketone with hydrogen gas in presence of nickel catalyst leads to the formation of a secondary alcohol.
[tex]{\text{4 - ethylhexan - 3 - one}}[/tex] reacts with hydrogen gas in the presence of nickel catalyst to undergo reduction reaction to form 4-ethyl-3-hexanol. (Refer to the attached image)
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Answer details:
Grade: High School
Subject: Chemistry
Chapter: Reduction Reaction
Keywords: Reduction reaction, hydrogen, addition, 4-ethyl-3-hexanol, aldehyde, ketone, double, single, nickel, catalyst, 4-ethylhexan-3-one, hydrogenation and 4-ethyl-3-hexanol.
Now consider lithium (LI+) and fluoride (F2) as oxidizing agents. How do these compare as oxidizing agents?
Answer:
Fluoride is a stronger oxidizing agent than lithium.
Explanation:
The element that has a valence configuration of 2s2 is ________.
Answer: The element having given valence electronic configuration is beryllium.
Explanation:
Electronic configuration is defined as the representation of electrons around the nucleus of an atom. Number of electrons in an atom are determined by the atomic number of that atom.
Valence electrons are defined as the electrons present in the outermost shell of an atom.
We are given:
Valence electronic configuration of an atom = [tex]2s^2[/tex]
So, the actual electronic configuration of atom will be [tex]1s^22s^2[/tex]
Total number of electrons = 2 + 2 = 4
So, the atomic number of given element is 4 and the element is beryllium
Hence, the element having given valence electronic configuration is beryllium.
How many moles of n are in 0.215 g of n2o?
Final answer:
To find the number of moles of nitrogen in 0.215 g of N2O, calculate the molar mass of N2O (44 g/mol), then use the mass to calculate the moles of N2O. Finally, multiply by 2 since there are two nitrogen atoms per molecule of N2O, resulting in approximately 0.00978 moles of nitrogen.
Explanation:
To calculate the number of moles of nitrogen in 0.215 g of N2O, we first need the molar mass of N2O. The formula N2O contains two nitrogen atoms (N) and one oxygen atom (O). Since N has a molar mass of 14 g/mol, and O has a molar mass of 16 g/mol, the molar mass of N2O is calculated as:
[tex](2 mol N) times (14 g/mol) + (1 mol O) times (16 g/mol) = 28 g/mol + 16 g/mol = 44 g/mol.[/tex]
Next, we use the formula:
Number of moles (n) = mass (m) / molar mass (M)
n = 0.215 g / 44 g/mol
Therefore, the moles of N2O in 0.215 g is calculated as approximately:
n = 0.00489 moles of N2O
To find the number of moles of nitrogen (N) in N2O, we need to remember that each molecule of N2O contains two atoms of nitrogen, so we multiply the moles of N2O by 2:
n(N) = 0.00489 moles of N2O
n(N) = 0.00978 moles of nitrogen
Two different compounds are obtained by combining nitrogen with oxygen. the first compound results from combining 46.7 g of n with 53.3 g of o, and the second compound results from combining 17.9 g of n and 82.0 g of o. calculate the ratio of the mass ratio of o to n in the second compound to the mass ratio of o to n in the first compound. express your answer numerically.
The mass ratio of oxygen to nitrogen in the first compound is 1.1413, and in the second compound, it is 4.5810. The ratio of these mass ratios is approximately 4.015.
Explanation:The question involves calculating the mass ratio of oxygen to nitrogen for two nitrogen-oxygen compounds and then finding the ratio of these mass ratios.
For the first compound:
Mass of O: 53.3 g
Mass of N: 46.7 g
Mass ratio of O to N: 53.3 g / 46.7 g = 1.1413
For the second compound:
Mass of O: 82.0 g
Mass of N: 17.9 g
Mass ratio of O to N: 82.0 g / 17.9 g = 4.5810
Now, we calculate the ratio of the mass ratio of the second compound to that of the first compound:
4.5810 / 1.1413 = 4.015
Therefore, the ratio of the mass ratio of oxygen to nitrogen in the second compound to the first one is approximately 4.015.