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:
1M HCl: add 1mol/12M = 83 ml conc. HCl to 1L of water or 8.3ml to 100ml.
2M HCl: add 2mol/12M = 167 ml conc. HCl to 1L of water or 16.7ml to 100ml.
You need to find which intermolecular forces are between the molecules
dipole-dipole,h bonds, etc.
I'm not very good at explaining but this is what my prof said to help us
Identify the class of the molecule or molecules you are given. Are they nonpolar species, ions or
do they have permanent dipoles? Is there only one species or are there two?
In the case of ONE species (i.e., a pure substance), the intermolecular forces will be between
molecules of the same type. So if you are dealing with ions, the intermolecular forces will be ION-
ION or IONIC. If you are dealing with dipoles, then the intermolecular forces will be DIPOLE-
DIPOLE. If you are dealing with nonpolar species, the intermolecular forces will be DISPERSION
or VAN DER WAALS or INDUCED DIPOLE-INDUCED DIPOLE (the last three are desciptions
of the same interaction; regrettably we cannot call them nonpolar-nonpolar!).
In the case of TWO species (i.e., a mixture), the intermolecular forces will be between molecules of
one type with molecules of the second type. For example, ION-DIPOLE interactions exist between
ions dissolved in a dipolar fluid such as water.
