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Semmy
11 days ago
7

An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an

d 8. When 75 mL of 0.1 M NaOH was added to 100mL of a 0.1 M solution of X at pH 2.0, the pH increased to 6.72. Calculate the pKa of the second group of X.
Source https://www.physicsforums.com/threads/calculating-pka.89490/
Chemistry
1 answer:
alisha [964]11 days ago
8 0

Answer:

7.3

Explanation:

Using the Henderson Hasselbalch equation, one can determine the pH or pOH of a solution via its pKa. Remember, pH = -log[H^{+}], and pKa = -logKa, where Ka denotes the acid's equilibrium constant.

The Henderson Hasselbalch formula:

pH = pKa - log \frac{[HA]}{[A^{-}]}

In this context, acid X possesses two ionic forms: the carboxyl group and an alternative form. Initially, we have 0.1 mol/L of acid in 100 mL, which gives:

n1 = (0.1 mol/L)×(0.1 L) = 0.01 mol

Upon dissociation, it yields 0.005 mol of the carboxyl form and 0.005 mol of the other form with stoichiometry assumed constant.

Introducing NaOH at a concentration of 0.1 mol/L and 75 mL, the moles of OH^- become:

n2 = (0.1 mol/L)×(0.075 L) = 0.0075 mol

Thus, 0.0075 mol of OH^- reacts with 0.005 mol of the carboxyl form, leading to 0.0025 mol of OH^-, which in turn reacts with 0.005 mol of the alternating group, leaving 0.0025 mol of the latter.

The new solution’s volume is 175 mL, but the concentrations of both forms remain unchanged in volume, so we can utilize the moles in the equation.

<pNote, the moles of the acid form remain 0.01 mol as it doesn’t undergo reaction!

Thus, we arrive at:

6.72 = pKa - log \frac{0.01}{0.0025}

6.72 = pKa - log 4

pKa - log 4 = 6.72

pKa = 6.72 + log 4

pKa = 6.72 + 0.6

pKa = 7.3

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A compound has a mass percentage of 53.46% C, 6.98% H, and 39.56% O. What is the empirical formula for this compound
lorasvet [946]

Answer:

The empirical formula is: C₂H₃O

Explanation:

The empirical formula, often referred to as the “minimum formula,” provides the simplest way to represent a chemical compound, illustrating both the present elements and their minimum integer ratios in terms of atoms.

The mass percentage composition indicates the proportion by mass of each element in a compound.

Using 100 g of the compound as a reference, the percentage can be expressed in grams. Thus, presuming 100 g of the compound consists of 53.46 g of C, 6.98 g of H, and 39.56 g of O.

Considering the molecular masses of each component, one can derive the number of relative atoms for each element:

C: 53.46 g *\frac{1 mol}{12.01 g } = 4.45 moles

H:6.98 g *\frac{1 mol}{1.01 g } = 6.91 moles

O:39.56 g *\frac{1 mol}{16g } = 2.47 moles

Subsequently, divide each value by the smallest one:

C: \frac{4.45 moles}{2.47 moles}=  1.80

H:\frac{6.91 moles}{2.47 moles}= 2.8

O:\frac{2.47 moles}{2.47 moles}=1

Approximating decimals to the nearest integer gives:

C: 2

H: 3

O: 1

Therefore the empirical formula is: C₂H₃O

5 1
13 days ago
A laboratory utilizes a mixture of 10% dimethyl sulfoxide (DMSO) in the freezing and long-term storage of embryonic stem cells.
Alekssandra [962]

Answer:

The right answer is "1.0100".

Explanation:

Assuming the total volume of the mixture is 100 ml.

Thus,

The volume of DMSO will be 10 mL and the volume of water will be 90 mL.

For DMSO:

= 10\times 1.1004

= 11.004 \ g

The total mass of the mixture will be:

= 90+11.004

= 101.004 \ g

Calculating the density of the mixture:

= \frac{Mass}{Volume}

= \frac{101.004}{100}

= 1.01004 \ g/mL

Thus,

The specific gravity of the mixture is:

= \frac{Density \ of \ mixture}{Density \ of \ water}

= \frac{1.01004}{1}

= 1.0100

3 0
6 days ago
Two glasses labeled A and B contain equal amounts of water at different temperatures. Kim put an antacid tablet into each of the
eduard [944]

Answer:

The correct statement is D. The water in Glass A is at a lower temperature compared to Glass B; consequently, the particles in Glass A exhibit reduced movement.

Explanation:

Raising the temperature increases the solubility of solutes.

The experiment indicates that's glass B is at a higher temperature than glass A since the antacid dissolves more quickly in glass B than in glass A. Therefore, glass A must be cooler, leading to slower particle movement compared to glass B.

6 0
10 days ago
Read 2 more answers
The graph above shows the changes in temperature recorded for the 2.00 l of h2o surrounding a constant-volume container in which
Alekssandra [962]

Answer:

25.2 kJ

Explanation:

The full question can be found in the image linked to this response.

It's important to highlight that the heat absorbed by the 2.00 L of water for increasing its temperature from the beginning to the end comes solely from the burning of benzoic acid, as there are no heat transfers to the container or the surroundings.

To find the heat released from benzoic acid combustion, we simply measure the heat needed to warm the water.

Q = mCΔT

To find the mass of the water,

Density = (mass)/(volume)

Mass = Density × volume

Density = 1 g/mL

Volume = 2.00 L = 2000 mL

Mass = 1 × 2000 = 2000 g

C = specific heat of water = 4.2 J/g.°C

ΔT = (final temperature) - (Initial temperature)

<pAccording to the graph,

Final water temperature = 25°C

Initial water temperature = 22°C

ΔT = 25 - 22 = 3°C

Q = (2000×4.2×3) = 25,200 J = 25.2 kJ

Hope this Helps!!!

6 0
8 days ago
Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Ca2+, V5+, Br-
lorasvet [946]

Answer:

CaS, CaBr₂, VBr₅, and V₂S₅.

Explanation:

  • The ionic compound must exhibit neutrality; its total charge should equal zero.
  • A binary ionic compound is formed from two distinct ions.

Ca²⁺ combines with either Br⁻ or S²⁻ to create binary ionic compounds.

  • CaS is created when Ca²⁺ pairs with S²⁻ resulting in the neutral binary ionic compound CaS.
  • CaBr₂ results from the combination of one mole of Ca²⁺ with two moles of Br⁻ to form the neutral binary ionic compound CaBr₂.

V⁵⁺ can also unite with either Br⁻ or S²⁻ to produce binary ionic compounds.

  • V₂S₅ is formed when two moles of V⁵⁺ bond with five moles of S²⁻ yielding the neutral binary ionic compound V₂S₅.
  • VBr₅ is produced by combining one mole of V⁵⁺ with five moles of Br⁻ to form the neutral binary ionic compound VBr₅.

Thus, the empirical formulas for four binary ionic compounds that may be produced are: CaS, CaBr₂, VBr₅, and V₂S₅.

5 0
12 days ago
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