1.)Find the PH of a solution with 0.20M NH3 with 0.15M of NH4Cl.
2.)Find PH of a solution with a mixture of 200mL of 0.50M HCHO2 and of 100mL of .50M NaCHO2.
Both of these are buffers. Use the Henderson-Hasselbalch equation.
1.) To find the pH of a solution with a mixture of NH3 and NH4Cl, we need to consider the acid-base equilibrium between NH3 (ammonia) and NH4+ (ammonium ion).
NH3 acts as a base and NH4+ acts as an acid. The equilibrium equation for this reaction is:
NH3 + H2O ⇌ NH4+ + OH-
First, we need to find the concentration of OH- ions. Since NH3 is a weak base, it undergoes partial ionization. The concentration of OH- ions can be calculated using the expression for the base ionization constant (Kb) of NH3.
The Kb for NH3 is given as 1.8 × 10^-5.
Using the balanced equation, we can see that the concentration of OH- ions is equal to the concentration of NH4+. Therefore, the concentration of OH- ions is 0.15 M.
Next, we can use the concentration of OH- ions to find the concentration of H+ ions. Since water is amphoteric, it can donate and accept protons. The product of [H+] and [OH-] is constant at 1.0 × 10^-14 in any aqueous solution.
[H+] × [OH-] = 1.0 × 10^-14
[H+] × 0.15 = 1.0 × 10^-14
[H+] = (1.0 × 10^-14) / 0.15
[H+] = 6.67 × 10^-14
Finally, we can use the concentration of H+ ions to find the pH using the formula:
pH = -log[H+]
pH = -log(6.67 x 10^-14)
pH ≈ 13.18
Therefore, the pH of the solution is approximately 13.18.
2.) To find the pH of a solution with a mixture of HCHO2 (formic acid) and NaCHO2 (sodium formate), we need to consider the acid-base equilibrium between HCHO2 and CHO2- (formate ion).
HCHO2 acts as an acid and CHO2- acts as a base. The equilibrium equation for this reaction is:
HCHO2 + H2O ⇌ H3O+ + CHO2-
First, we need to find the concentration of H3O+ ions. Since HCHO2 is a weak acid, it undergoes partial ionization. The concentration of H3O+ ions can be calculated using the expression for the acid ionization constant (Ka) of HCHO2.
The Ka for HCHO2 is given as 1.8 × 10^-4.
Using the balanced equation, we can see that the concentration of H3O+ ions is equal to the concentration of HCHO2. Therefore, the concentration of H3O+ ions is 0.50 M.
Next, we can use the concentration of H3O+ ions to find the pH using the formula:
pH = -log[H3O+]
pH = -log(0.50)
pH ≈ 0.30
Therefore, the pH of the solution is approximately 0.30.
To find the pH of a solution, you need to determine the concentration of hydrogen ions (H+). There are multiple methods you can use, such as using the concentration of a strong acid or the dissociation of a weak acid/base.
1.) Finding the pH of a solution with 0.20M NH3 and 0.15M NH4Cl:
In this case, NH3 acts as a base, and NH4Cl is its conjugate acid.
Step 1: Write the balanced equation for the reaction between NH3 and H2O:
NH3 + H2O ⇌ NH4+ + OH-
Step 2: Identify that NH4+ is the conjugate acid and NH3 is the weak base.
Step 3: Set up an ICE (Initial, Change, Equilibrium) table to track the reaction's progress. Assume x as the amount of NH4+ that dissociates.
NH3 + H2O ⇌ NH4+ + OH-
Initial: 0.20M 0.00M 0.00M 0.00M
Change: -x +x +x +x
Equilibrium: 0.20M-x x x x
Step 4: Write the expression for the equilibrium constant (Kb) for the reaction:
Kb = ([NH4+][OH-]) / ([NH3])
Step 5: Substitute the equilibrium concentrations from the ICE table into the Kb expression:
Kb = (x * x) / (0.20M - x) = 1.8 x 10^-5 (substitute the value of Kb for NH3)
Step 6: Solve the quadratic equation for x. Since the Kb for NH3 is small, we can assume that (0.20M - x) is approximately 0.20M and simplify the equation:
(x * x) / 0.20M = 1.8 x 10^-5
x * x = (1.8 x 10^-5) * 0.20M
x^2 = 3.6 x 10^-6
x = 1.9 x 10^-3 (approx)
Step 7: Calculate the OH- concentration:
OH- = x = 1.9 x 10^-3 M
Step 8: Calculate the pH:
pOH = -log10(OH-) = -log10(1.9 x 10^-3) ≈ 2.72
Since pOH + pH = 14, pH ≈ 11.28
Therefore, the pH of the solution with 0.20M NH3 and 0.15M NH4Cl is approximately 11.28.
2.) Finding the pH of a solution with a mixture of 200mL of 0.50M HCHO2 and 100mL of 0.50M NaCHO2:
In this case, HCHO2 acts as a weak acid, and NaCHO2 is its conjugate base.
Step 1: Write the balanced equation for the reaction between HCHO2 and H2O:
HCHO2 + H2O ⇌ H3O+ + CHO2-
Step 2: Identify that HCHO2 is the weak acid and CHO2- is the conjugate base.
Step 3: Calculate the moles of HCHO2 and NaCHO2:
moles of HCHO2 = (0.50M) * (0.200L) = 0.10 moles
moles of NaCHO2 = (0.50M) * (0.100L) = 0.05 moles
Step 4: Calculate the total volume of the solution:
total volume = 200mL + 100mL = 300mL = 0.3L
Step 5: Calculate the combined concentration of HCHO2 and CHO2- in the mixture:
[HCHO2] = (0.10 moles) / (0.3L) = 0.33M
[CHO2-] = (0.05 moles) / (0.3L) = 0.17M
Step 6: Write the expression for the equilibrium constant (Ka) for the reaction:
Ka = ([H3O+][CHO2-]) / ([HCHO2])
Step 7: Substitute the equilibrium concentrations into the Ka expression:
Ka = ([H3O+][0.17M]) / (0.33M) = 1.8 x 10^-4 (substitute the value of Ka for HCHO2)
Step 8: Solve for [H3O+]:
[H3O+] = (Ka * [HCHO2]) / [CHO2-] = (1.8 x 10^-4 * 0.33M) / 0.17M ≈ 3.5 x 10^-4 M
Step 9: Calculate the pH:
pH = -log10([H3O+]) = -log10(3.5 x 10^-4) ≈ 3.46
Therefore, the pH of the solution with a mixture of 200mL of 0.50M HCHO2 and 100mL of 0.50M NaCHO2 is approximately 3.46.