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You are given sodium acetate, 1M HCl, NaHCO3 and Na2CO3. Determine which of these four you would need and then show calculations to make buffer pH = 4.7 by method 2. Assume making 100 mls of a 0.1 M buffer. Neutralization of a portion of the weak base with sufficient strong acid to give the desired ratio, [A- ]/[HA]: Na+ + A- + HCl → HA + Na+ + Cl- I've been having a lot of trouble going step by step with this one. If you could really explain the logic behind each step it would be very much appreciated!! I'm also confused about the ka value. Am I just supposed to look that up? Thank you so much for your time!

2 months ago

Solution 1

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2 months ago

Answer:

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Explanation:

📚 Related Questions

Question

Hard liquors or spirits, such as brandy, whisky, and tequila, are commonly produced by fermentation of carbohydrate-rich natural products. In the process, a mixture containing water (C), ethanol (B), methanol (A), and many other components is generated. Methanol needs to be removed because of its toxic properties; water is extracted to produce beverages with various concentrations of ethanol. a. What differentiating characteristic could be used to separate methanol, ethanol, and water? What experimental technique could be used to separate these substances? b. As you heat up the alcoholic mixture, in which order will the three substances (A-Methanol, B-Ethanol, C-Water) separate? c. At which temperatures will each fraction distill? d. Is there any advantage in changing the pressure at which the distillation is performed? Why?

Solution 1

Answer:

a. differences in boiling points, fractional distillations,water,ethanol and methanol, 100 ,78,56, no

Explanation:

alcohol chemistry

Question

In the reaction 3H2 + N2 → ____ 2 NH3, what coefficient should be placed in front of NH3 to balance the reaction? A) 1 B) 2 C) 3 D) no coefficient is needed

Solution 1

Answer:

In the reaction 3H2 + N2 → ____ 2 NH3

D) no coefficient is needed

as it is balanced already.

Solution 2

Answer:

Explanation:

Question

The density of benzene at 15 ∘C∘C is 0.8787 g/mg/mL. Calculate the mass of 0.1700 LL of benzene at this temperature.

Solution 1

Answer:15 gl

Explanation:

Question

What is the specific volume of oxygen at 25.5 psia and 80°F? The gas constant of oxygen is R = 0.3353 psia·ft3/lbm·R.

Solution 1

Specific volume: $3.9 ft^3/mol = 0.109 m^3/mol$

Explanation:

We can solve this problem by applying the equation of state for ideal gases, which states that:

$pV=nRT$

where

p is the gas pressure

V is the gas volume

n is the number of moles

R is the gas constant

T is the absolute temperature

The equation can be re-arranged as

$V=\frac{nRT}{p}$

Here we want to find the specific volume, which is the volume of gas per number of moles, so we have to divide by n:

$V_s=\frac{RT}{p}$

For the gas in this problem, we have:

$p=25.5 psi$ is the pressure

$T=80F$ , converting to Kelvin:

$T=(80-32)\cdot \frac{5}{9}+273.15=299.8 K$

$R=0.3353 psi\cdot ft^3 /(lbmR)$ is the gas constant

Solving for Vs,

$V_s=\frac{(0.3353)(299.8)}{25.5}=3.9 ft^3$ /mol

And since

1 feet = 0.3048 m

$(1 feet)^3 = (0.3048 m)^3 = 0.028 m^3$

Then the specific volume in SI units is

$V_s = 3.9 ft^3 \cdot 0.028 = 0.109 m^3$ /mol

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Question

Write the formula of the conjugate base for the following acids: Acid Conjugate base (a) HNO2 (b) H2SO4 (c) H2S (d) HCN (e) HCOOH

Solution 1

The conjugate bases of the following acids are as follows:

HNO2 = NO2-
H2SO4 = HSO4-
H2S = HS-
HCN = CN-
HCOOH = COOH-

CONJUGATE BASE:

According to Bronsted-Lowry definition, a conjugate base is formed when a proton is removed from an acid while a conjugate acid is formed when a proton is added to a base.

An acid is known to donate a proton when dissolved in an aqueous solution, the resulting anion formed is called a conjugate base.

The conjugate bases formed by the following acids when they lose a proton is as follows:

HNO2 = NO2-
H2SO4 = HSO4-
H2S = HS-
HCN = CN-
HCOOH = COOH-

Learn more at: brainly.com/question/12883745?referrer=searchResults

Solution 2

Answer:

Answers are in the examples of Conjugate Base.

Explanation:

Conjugate Acid is a specie which is formed when a base abstract or picks a proton (H⁺).

Example:

NH₃ + HCl → NH₄⁺ + Cl⁻

In this example Ammonia (NH₃) is a base which after reacting with acid forms the conjugate acid (NH₄⁺).

Conjugate Base is a specie which is formed when an acid looses / donates its proton (H⁺) to a base.

Example:

(a)

B + HNO₃ → BH⁺ + NO₃⁻

In this example HNO₃ (Nitric Acid) is the acid which on reaction with base forms the conjugate base ( NO₃⁻) called nitrate.

(b)

B + H₂SO₄ → BH⁺ + HSO₄⁻

B + HSO₄⁻ → BH⁺ + SO₄⁻

In this example H₂SO₄ (Sulfuric Acid) is the acid which on reaction with base forms the conjugate bases ( HSO₄⁻ and SO₄⁻) respectively.

(c)

B + H₂S → BH⁺ + HS⁻

In this example H₂S (Hydrogen Sulfide) is the acid which on reaction with base forms the conjugate base HS⁻.

(c)

B + HCOOH → BH⁺ + HCOO⁻

In this example HCOOH (Formic Acid) is the acid which on reaction with base forms the conjugate base HCOO⁻ called formate ion.

Question

Should you report a melting point as a single number or a range? How are the high and low endpoints of the range determined experimentally during the melting point experiment? What is the relationship between the purity of a compound and its melting point range?

Solution 1

The melting range results from the fact that different components of the impure material melt at different temperatures. Thermodynamically, the material minimizes its Gibbs free energy (ie maximizes its entropy) during this process, which determines the phase mixtures with the lowest energy states.

During the melting point test, when the first crystals of the compound begin to melt, mark this as the lower limit (low end point). When all the crystals have melted, mark this as the upper limit (high point).

Even a small amount of impurities lowers the melting point of a compound by a few degrees and extends the temperature range of the melting point. Because the impurity causes defects in the crystal lattice, it is easier to overcome intermolecular interactions.

What is melting point?

Melting point is the temperature at which a certain solid changes from a solid state to a liquid or melts. The melting point of a substance is the temperature at which it changes its state from solid to liquid. At the melting point, the solid and liquid phases are in equilibrium. The melting point of a substance depends on pressure and is usually defined at a constant pressure, such as 1 atmosphere or 100 kPa. Molecular composition, attraction and impurities can all affect the melting point of substances.

To learn more about melting point, refer;

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Solution 2

Answer:

a pure substance has a specific melting point but a non-pure substance has a its melting point over a range

Explanation:

test of purity

Question

To what volume should you dilute 60.0 mL of a 4.50 M KI solution so that 30.0 mL of the diluted solution contains 3.25 g of KI?

Solution 1

Answer:

You should dilute to 414.1 mL

Explanation:

Let's apply the dilution formula to solve this:

M conc . Vol conc = M dil . Vol dil

Molarity for concentrated solution → 4.50 M

Volume for concentrated solution → 60 mL

Molarity for diluted solution → Unknown

Volume for diluted solution → Answer

Let's determine the molarity for diluted solution with data given.

We convert the mass of KI to moles

3.25 g . 1 mol/166 g = 0.0195 moles

Let's convert the volume from mL to L

30 mL . 1 L/1000mL = 0.030 L

Molarity (mol/L) = 0.0195 mol /0.030L = 0.652 M

We replace the data in the formula

4.50 M . 60 mL = 0.652 M . Volume for diluted solution

(4.50M . 60mL) / 0.652M = Volume for diluted solution

414.1 mL = Volume for diluted solution

Solution 2

Answer: 413mL

Explanation:

Molar Mass of KI = 39 + 127 = 166g/mol

Mass conc of KI = 3.25g

Number of mole = Mass conc/Molar Mass

Number of mole = 3.25/166 = 0.0196 mol

But recall

Molarity = number of mole / Volume

Number of mole = 0.0196 mol

Volume = 30mL = 30/1000 = 0.03L

Molarity = 0.0196/0.03

Molarity = 0.653M

C1 = 4.5M

V1 = 60mL = 60/1000 = 0.06L

C2 = 0.653M

V2 =?

C1V1 = C2V2

4.5 x 0.06 = 0.653 x V2

V2 = (4.5 x 0.06) / 0.653

V2 = 0.413L

Converting to mL

0.413 x 1000 = 413mL

Question

A 10 mL sample of HCl solution was transferred by pipet to an Erlenmeyer flask and then diluted by adding about 40 mL of distelled water. What is the approximate H3O+ concentration and pH of the solution in the flask before the titration begins?

Solution 1

Answer: (a) 0.1M

(b) pH = 1

Explanation:

Moles = Molarity × Volume me(L)

The initial number of moles in the flask = 0.5M × 0.010L = 0.005moles

After the addition of 40 mL of in the flask again, the total volume in the flask = 10 mL + 40 mL = 50 mL

Concentration = Moles/Volume(L)[H3O+]

0.005moles/0.050L = 0.1M

pH = −log[H3O+] = −log(0.1) = 1

Solution 2

Answer:

The approximate H3O+ concentration before the titration begins is 0.1 M

pH of the solution in the flask is 1

Explanation:

Initial concentration of HCl = 0.5 mol/L

Initial number of mols in Erlenmeyer flask = (0.5 mol/L)(0.010 L) = 0.005 mol

After addition of 40 mL of volume to the flask again, the total volume in the flask = 50 mL

Therefore - the resultant [H3O+] concetration in the solution = 0.005 mol / 0.050 L = 0.1 M (before titration begins)

Hence - pH = -log[H3O+] = -log (0.1) = 1

Question

Molybdenum (Mo) has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Compute its density.

Solution 1

Answer:

The density of the crystal is 10.22 g/cm³

Explanation:

Step 1: Data given

Molybdenum (Mo) has a BCC crystal structure

atomic radius = 0.1363 nm

atomic weight of 95.94 g/mol

Step 2: Calculate density of a bcc crystal

Density ρ= (nA *Mo)/(Vc*Na)

⇒For BCC, n = 2 atoms/unit cell, and realizing that V c = a ³

a = 4R/√3

Vc =(4R/√3)³

AMo = atomic weight = 95.94 g/mol

ρ = (n*A Mo)/((4R/√3)³ *Na)

⇒with n = 2 atoms/unit cell

⇒ with AMo = 95.94 g/mol

⇒ with Vc =(4R/√3)³ = (4*0.1363*10-7 cm)³ /(√3)³

⇒with Na = 6.022*10^23 atoms/mol

ρ =10.22 g/cm³

The density of the crystal is 10.22 g/cm³

Solution 2

Answer:

The density of the molybdenum BCC crystal structure is 10.22 g/cm^3

Explanation:

Density (D) of BCC crystal structure is given as:

D = (n × AW) ÷ (12.32r^3 × Na)

n is the number of atoms of molybdenum per unit cell = 2

AW is the atomic weight of molybdenum = 95.94 g/mol

r is the atomic radius of molybdenum = 0.1363 nm = 0.1363×10^-9 m = 0.1363×10^-9 × 100 = 1.363×10^-8 cm

Na is Avogadro's number = 6.022×10^23 atoms/mol

D = (2×95.94) ÷ [12.32 × (1.363×10^-8)^3 × 6.022×10^23] = 191.88 ÷ 18.78 = 10.22 g/cm^3

Question

Calculate the composition, in weight percent, of an alloy that contains 218.0 kg titanium, 14.6 kg of aluminum, and 9.7 kg of vanadium.

Solution 1

Answer:

Composition of Titanium in weight percent is 89.97 wt%

Composition of Aluminum in weight percent is 6.03 wt%

Composition of Vanadium in weight percent is 4.00 wt%

Explanation:

To solve this problem, we use the following formula

$C_{Ti} = \frac{M_{Ti}}{M_{Ti} + M_{Al} + M_{V}} X 100$

where;

$C_{Ti}$ is composition of Titanium in percentage

$M_{Ti}$ is mass of titanium

$M_{Al}$ is mass of aluminum

$M_V$ is mass of vanadium

Step (1) : Composition of Titanium in weight percent

$C_{Ti} = \frac{M_{Ti}}{M_{Ti} + M_{Al} + M_{V}} X 100$

= (100*218)/(218+14.6+9.7)

= 89.97 wt%

Step (2) : Composition of Aluminum in weight percent

$C_{Al} = \frac{M_{Al}}{M_{Ti} + M_{Al} + M_{V}} X 100$

= (100*14.6)/(218+14.6+9.7)

= 6.03 wt%

Step (2) : Composition of Vanadium in weight percent

$C_{V} = \frac{M_{V}}{M_{Ti} + M_{Al} + M_{V}} X 100$

= (100*9.7)/(218+14.6+9.7)

= 4.00 wt%

Solution 2

Answer:

% titanium = 89.97 %

% aluminium = 6.03 %

% vanadium = 4.00 %

Explanation:

Step 1: Data given

An alloy contains 218.0 kg titanium, 14.6 kg of aluminium and 9.7 kg of vanadium

Step 2: Calculate total mass

Total mass = mass titanium + mass aluminium + lass vanadium

Total mass = 218.0 + 14.6 + 9.7

Total mass = 242.3 kg

Step 3: Calculate weight percent

Weight percent = (mass / total mass) * 100%

% titanium = (218.0 kg / 242.3 kg)*100%

% titanium = 89.97 %

% aluminium = (14.6 kg / 242.3 kg) * 100 %

% aluminium = 6.03 %

% vanadium = (9.7 kg/ 242.3 kg) *100 %

% vanadium = 4.00 %

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