Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
Answer:
Relative and average atomic mass both describe properties of an element related to its different isotopes.
Explanation:However, relative atomic mass is a standardized number that's assumed to be correct under most circumstances, while average atomic mass is only true for a specific sample.
Answer:
P₂ = 28.5 torr
Explanation:
Given data:
Initial pressure = 38 torr
Initial volume = 500 L
Final volume = 677 L
Final pressure = ?
Solution:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = Initial volume
P₂ = Final pressure
V₂ = Final volume
Now we will put the vales in formula.
P₁V₁ = P₂V₂
P₂ = P₁V₁ /V₂
P₂ = 38 torr × 500 L / 667 L
P₂ = 19000 torr. L / 667 L
P₂ = 28.5 torr
Answer is: intramolecular attractions are stronger.
Intramolecular attractions are the forces between atoms in molecule.
There are several types of intramolecular forces: covalent bonds, ionic bonds.
Intermolecular forces are the forces between molecules. The stronger are intermolecular forces, the higher is boiling point of compound, because more energy is needed to break interaction between molecules.
There are several types of intermolecular forces: hydrogen bonding, ion-induced dipole forces, ion-dipole forces andvan der Waals forces.
Hydrogen bonds are approximately 5% of the bond strength of covalent C-C or C-H bonds.
Hydrogen bonds strength in water is approximately 20 kJ/mol, strenght of carbon-carbon bond is approximately 350 kJ/mol and strengh of carbon-hydrogen bond is approximately 340 kJ/mol.
20 kJ/350 kJ = 0.057 = 5.7 %.
Answer:
In pair NaF and H2O both compounds exibit predominantly ionic bonding.
