Calculate the pH of (1)0.150M KC2H3O2:
(2)0.200M pyridine chloride, C5H5NHCl: and(3) 0.500M NaCl
I'll do the first one. The others are the same type although the reactions are different.
KC2H3O2--potassium acetate is a salt. The acetate ion is a base and it hydrolyzes in water.
C2H3O2^- + HOH ==>HC2H3O2 + OH^-
Set up an ICE chart and substitute into the equilibrium equation below as follows:
Kb = (Kw/Ka) = (HC2H3O2)(OH^-)/(C2H3O2^-)
You know Kw and Ka. (HC2H3O2) = x = (OH^-) and (C2H3O2^-) = 0.150-x
Solve for x which is (OH^-), convert to pOH and from there to pH.
(2)C5H5NHCl is a salt. The C5H5NH^+ is an acid.
C5H5NH^+ + H2O ==> H3O^+ + C5H5N
(3)NaCl is a salt. Neither the Na^+ nor the Cl^- hydrolyze; therefore, the pH is that of pure water.
H2O ==> H^+ + OH^-
Kw = (H^+)(OH^-) = 1E-14
To calculate the pH of these solutions, we need to determine whether the substances dissociate in water and if they act as acids or bases.
(1) 0.150M KC2H3O2:
Potassium acetate (KC2H3O2) is a salt formed by the reaction of a weak acid (acetic acid, CH3COOH) and a strong base (potassium hydroxide, KOH). In water, KC2H3O2 dissociates into its ions: K+ and C2H3O2-.
C2H3O2- (acetate ion) is the conjugate base of acetic acid, which is a weak acid. Therefore, it undergoes hydrolysis in water, producing hydroxide ions (OH-) and acetic acid (CH3COOH).
Since hydroxide ions (OH-) are produced, KC2H3O2 acts as a base. To determine the pH of the solution, we need to calculate the concentration of hydroxide ions (OH-).
To find the concentration of OH-, we can use the equation for the hydrolysis reaction of the acetate ion:
C2H3O2- + H2O ⇌ CH3COOH + OH-
To calculate the concentration of OH-, we can assume that x is the concentration of OH- formed and neglect the x in comparison to 0.150 (initial concentration of KC2H3O2) to simplify the calculation.
Therefore, the concentration of OH- can be approximated as 0.150M × (x/0.150), where x represents the concentration of OH- ions formed.
However, we need to take into account that acetic acid (CH3COOH) is a weak acid that partially dissociates, so the concentration of OH- will be slightly lower.
To simplify, we can assume that the concentration of OH- formed is equal to x. This approximation will give us a reasonable estimate of the pH of the solution.
Next, we can use the equation for the water dissociation constant (Kw) to find the concentration of hydrogen ions (H+) in the solution:
Kw = [H+][OH-] = 1.0 × 10^-14 M^2
Given that the concentration of OH- is x, we can set up the equation:
(1.0 × 10^-14) = (x)(x)
Solving this equation gives us x ≈ 1.0 × 10^-7 M as the concentration of OH- ions.
Since the solution is basic, the concentration of H+ ions will be lower than 1.0 × 10^-7 M.
Therefore, the pH of the 0.150M KC2H3O2 solution will be approximately equal to 14 minus the logarithm (base 10) of the concentration of H+ ions:
pH ≈ 14 - log[H+]
pH ≈ 14 - log[1.0 × 10^-7]
pH ≈ 14 + 7
pH ≈ 11
Therefore, the pH of the 0.150M KC2H3O2 solution is approximately 11.
(2) 0.200M pyridine chloride, C5H5NHCl:
Pyridine chloride (C5H5NHCl) is a salt formed by the reaction of a weak base (pyridine, C5H5N) and a strong acid (hydrochloric acid, HCl). In water, C5H5NHCl dissociates into its ions: C5H5N and HCl.
HCl is a strong acid that dissociates completely in water, producing H+ ions. Therefore, the concentration of H+ ions in the solution is equal to the initial concentration of HCl.
The pH of the solution can be calculated using the concentration of H+ ions:
pH = -log[H+]
pH = -log[0.200M]
pH ≈ -log(2.0 × 10^-1)
pH ≈ -(-0.70)
pH ≈ 0.70
Therefore, the pH of the 0.200M pyridine chloride solution is approximately 0.70.
(3) 0.500M NaCl:
Sodium chloride (NaCl) is a salt formed by the reaction of a strong acid (hydrochloric acid, HCl) and a strong base (sodium hydroxide, NaOH). In water, NaCl dissociates into its ions: Na+ and Cl-.
Neither Na+ nor Cl- ions react with water to produce H+ or OH- ions. Therefore, the solution of NaCl will be neutral, and the pH will be 7.
Therefore, the pH of the 0.500M NaCl solution is 7.
To calculate the pH of a solution, we need to consider the dissociation of the solute in water and the concentration of the resulting ions. Let's calculate the pH for each of the three solutions:
(1) 0.150M KC2H3O2:
KC2H3O2 is a salt that dissociates into K+ and C2H3O2- ions in water. The C2H3O2- ion can act as a weak base by accepting a proton from water. We need to consider the dissociation of the C2H3O2- ion, also known as the acetate ion.
C2H3O2- + H2O <--> HC2H3O2 + OH-
Since C2H3O2- is a weak base, we can simplify the equation to:
C2H3O2- <--> HC2H3O2 + OH-
The concentration of OH- ions is negligible compared to the concentration of C2H3O2- ions, so we can disregard it. This means that the concentration of OH- is equal to the concentration of HC2H3O2 (since one mole of C2H3O2- will react with one mole of water to form one mole of HC2H3O2).
Therefore, the concentration of OH- in the solution is 0.150M.
To calculate the pOH, we use the equation:
pOH = -log[OH-]
pOH = -log(0.150)
pOH ≈ 0.82
Since the solution is neutral (neither acidic nor basic), the pH is:
pH = 14 - pOH
pH = 14 - 0.82
pH ≈ 13.18
So, the pH of a 0.150M KC2H3O2 solution is approximately 13.18.
Next, let's calculate the pH for the other two solutions.
(2) 0.200M pyridine chloride, C5H5NHCl:
In this case, we have a salt that dissociates into the pyridine cation (C5H5NH+) and chloride anion (Cl-) in water. Neither of these ions significantly interacts with water, and therefore, does not affect the pH of the resulting solution. So, the pH of the 0.200M pyridine chloride solution is simply 7 (neutral).
(3) 0.500M NaCl:
NaCl is a strong electrolyte that dissociates completely into Na+ and Cl- ions in water. Again, these ions do not significantly interact with water, so their presence won't affect the pH. Hence, the pH of a 0.500M NaCl solution is also 7 (neutral).
To summarize:
- The pH of a 0.150M KC2H3O2 solution is approximately 13.18 (basic).
- The pH of a 0.200M pyridine chloride solution is 7 (neutral).
- The pH of a 0.500M NaCl solution is 7 (neutral).