A chemist titrates 90.0 mL of a 0.5870 M acetic acid (HCH, CO) solution with 0.4794M NaOH solution at 25 °C. Calculate the pH at equivalence. The p Kg of acetic acid is 4.76. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added

Answer:

See Explanation

Explanation:

Pb2O3 is better formulated as PbO.PbO2. It is actually a mixture of the two oxides of lead, lead II oxide and lead IV oxide.

This implies that this compound Pb2O3  (sometimes called lead sesquioxide) is a mixture of the oxides of lead in its two known oxidation states +II and +IV.

Hence Pb2O3  contains PbO and PbO2 units.

Answer:

The strength of a bond depends on the amount of overlap between the two orbitals of the bonding atoms

Orbitals bond in the directions in which they protrude or point to obtain maximum overlap

Explanation:

The valence bond theory was proposed by Linus Pauling. Compounds are firmed by overlap of atomic orbitals to attain a favourable overlap integral. The better the overlap integral (extent of overlap) the better or stringer the covalent bond.

Orbitals overlap in directions which ensure a maximum overlap of atomic orbitals in the covalent bond.

Answer:

THE STRENGTH OF THE BOND DEPENDS ON THE AMOUNT OF OVERLAP BETWEEN THE TWO ORBITALS OF THE BONDING ATOMS

ORBITALS BOND IN THE DIRECTION OR POINT IN WHICH THEY PROTRUDE OR POINT TO OBTAIN MAXIMUM OVERLAP.

Explanation:

Valence bond theory describes the covalent bond as the overlap of half-filled atomic orbital yields a pair of electrons shared between the two bonded atoms. Overlapping of orbitals occurs when a portion of one orbital and the other occur in the same region of space. The strength of a bond is determined by the amount of overlap between the two orbitals of the bonding atoms. In other words, orbitals that overlap more and in the right orientation of maximum overlapping form stronger bonds that those with less overlap and right orientation for maximum overlap. The bonding occurs at a varying distance in different atoms from which it obtains its stable energy caused by the increase in the attraction of nuclei for the electrons.

Orbitals also bond in the direction to obtain maximum overlap as orientation of the atoms also affect overlap. The greater overlap occurs when atoms are oriented on a direct line mostly end to end or side by side between the two nuclei depending on the type of bond formed. A sigma bond is formed when atoms overlap end to end in which a straight line exists between the two atoms that is the internuclear axis indicating the concentrated energy  density in that region. Pi bond exits in when overlap occurs in the side -to -side orientation and the energy density is concentrated opposite the internuclear axis.

Answer:

The molar solubility of NiS is 7.7 * 10⁻⁷ M

Explanation:

To answer this question, we need to keep in mind two equilibriums.

First, we have the solubilization of NiS:

NiS ⇄ Ni²⁺ + S²⁻   ksp= 3.0 * 10⁻²¹  (we know this from standard tables)

Second, we have the formation of the complex:

Ni²⁺ + 6NH₃ ⇄ [Ni(NH₃)₆]²⁺   kf=2.0 * 10⁻⁸

Combine the two equilibriums and we have

NiS + 6NH₃ ⇄ [Ni(NH₃)₆]²⁺ + S²⁻  K= ksp * kf =6.0* 10⁻¹³=

The molar solubility s is equal to both [Ni(NH₃)₆²⁺] and [S²]

At equilibrium, [NH₃]= 3,1 M - 6s

Thus, if we replace those terms in the formula for K, we're left with:

Using an approximation we can ignore the denominator and we have

• s²=6.0 * 10⁻¹³

• s=7.7 * 10⁻⁷

Answer: -

Many drugs are sold as their hydrochloric salts (RNH₃⁺Cl⁻), formed by reaction of an amine (RNH₂) with HCl.

It is done because amines are generally liquids. But their hydrochloric salts are solid. A solid drug is always more preferable for drug companies as their handling and packaging are easier.

Acebutolol consists of one amide functionality as well as a secondary amine functionality.

When HCl is added, the lone pairs of the nitrogen of the secondary amine attacks it, leading to the formation of it's hydrochloric salts.

Answer:

1. pH = 2,82

2. 3,20mL of 1,135M NaOH

3. pH = 3,25

Explanation:

The buffer of acetic acid (HC₂H₃O₂) is:

HC₂H₃O₂ ⇄ H⁺ + C₂H₃O₂⁻

The reaction of HC₂H₃O₂ with NaOH produce:

HC₂H₃O₂ + NaOH → C₂H₃O₂⁻ + Na⁺ + H₂O

And ka is defined as:

ka = [H⁺] [C₂H₃O₂⁻] / [HC₂H₃O₂] = 1,8x10⁻⁵ (1)

1. When in the solution you have just 0,13M HC₂H₃O₂ the concentrations in equilibrium will be:

[H⁺] = x

[C₂H₃O₂⁻] = x

[HC₂H₃O₂] = 0,13 - x

Replacing in (1)

[x] [x] / [0,13-x] = 1,8x10⁻⁵

x² = 2,34x10⁻⁶ - 1,8x10⁻⁵x

x² - 2,34x10⁻⁶ + 1,8x10⁻⁵x  = 0

Solving for x:

x = - 0,0015 (Wrong answer, there is no negative concentrations)

x = 0,0015

As [H⁺] = x = 0,0015 and pH is -log [H⁺], pH of the solution is 2,82

2. The equivalence point is reached when moles of HC₂H₃O₂ are equal to moles of NaOH. Moles of HC₂H₃O₂ are:

0,0466L × (0,078mol / L) = 3,63x10⁻³ moles of HC₂H₃O₂

In a 1,135M NaOH, these moles are reached with the addition of:

3,63x10⁻³ moles × (L / 1,135mol) = 3,20x10⁻³L = 3,20mL of 1,135M NaOH

3. The initial moles of HC₂H₃O₂ are:

0,0172L × (0,128mol / L) = 2,20x10⁻³ moles of HC₂H₃O₂

As the addition of NaOH spent HC₂H₃O₂ producing C₂H₃O₂⁻. Moles of C₂H₃O₂⁻ are equal to moles of NaOH and moles of HC₂H₃O₂ are initial moles - moles of NaOH. That means:

0,46x10⁻³L NaOH × (0,155mol / L) = 7,13x10⁻⁵ moles of NaOH ≡ moles of C₂H₃O₂⁻

Final moles of HC₂H₃O₂ are:

2,20x10⁻³ - 7,13x10⁻⁵ = 2,2187x10⁻³ moles of HC₂H₃O₂

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [C₂H₃O₂⁻] / [HC₂H₃O₂]

Where pka is -log ka = 4,74. Replacing:

pH = 4,74 + log₁₀ [7,13x10⁻⁵] / [2,2187x10⁻³ ]

pH = 3,25

I hope it helps!

Answer:

a) Precision

b) Accuracy

c) Accuracy and precision

d) Accuracy

Explanation:

When an experiment is done more than once to determine if the results are statistically ok, two forms of the validations are possible the accuracy and precision. When the values of the various experiments are close to the known value, then they are accurate. When the values are close to each other they are precise. So, sometimes the results are precise but are not accurate, and vice-versa.

a) Here, the person wants to find if the 5 aliquots will have close results, so, he or she is looking for precision.

b) Here the amount of analyte is already known, and the person wants to identify if the value will be the same when analyzed together with another sample, thus he or she is looking for accuracy.

c) Here the three results will be compared with each other (precision) and with the standard value of the method (accuracy).

d) The methods will be tested, and the values will be compared with the standard known value, so the person is looking for accuracy.