Calculate the pH of each of the following solutions.
(a) 0.300 M propanoic acid (HC3H5O2, Ka = 1.3 10-5)
(b) 0.300 M sodium propanoate (NaC3H5O2)
(c) pure H2O
(d) 0.300 M HC3H5O2 and 0.300 M NaC3H5O2
a)Let HA = the weak acid, proanoic acid.
..............HA ==> H^+ + A^-
initial......0.300....0......0
change.........-x......x......x
equil........0.300-x...x.......x
Ka = (H^+)(A^-)/(HA)
Substitute from the ICE chart and solve for x, then convert to pH.
b)
This is a salt and the pH will be determined by the hydrolysis of the salt.
A^- + HOH ==> HA + OH^-
Set up an ICE chart similar to the one in (a) and solve for OH^-, then convert to pH.
(c) Pure water ionizes as
HOH => H^+ + OH^-
Set up an ICE chart and use
Kw = (H^+)(OH^-), solve for H^+ and convert to pH.
(d)
This is a buffered solution. Use the Henderson-Hasselbalch equation.
Post your work if you get stuck.
To calculate the pH of each solution, you need to understand the concepts of acids, bases, and pH. The pH of a solution tells you how acidic or basic it is. pH is defined as the negative logarithm (base 10) of the concentration of hydrogen ions (H+) in a solution. The equation for calculating pH is: pH = -log[H+].
(a) Calculation for 0.300 M propanoic acid (HC3H5O2, Ka = 1.3 x 10^-5):
1. First, determine the concentration of H+ ions in the solution. From the balanced equation for the dissociation of propanoic acid, one mole of propanoic acid ionizes to produce one mole of H+ ions:
HC3H5O2 (aq) β H+ (aq) + C3H5O2- (aq)
2. Use the equation for the dissociation constant (Ka) to find the concentration of H+ ions:
Ka = [H+][C3H5O2-]/[HC3H5O2]
1.3 x 10^-5 = [H+][C3H5O2-]/0.300
3. Since the concentration of propanoic acid (HC3H5O2) is 0.300 M, the concentration of H+ ions is equal to the initial concentration of propanoic acid that dissociates, assuming a complete dissociation:
[H+] = [HC3H5O2] = 0.300 M
4. Plug the value of [H+] into the pH equation:
pH = -log[(0.300)]
pH β 0.522
The pH of a 0.300 M propanoic acid solution is approximately 0.522.
(b) Calculation for 0.300 M sodium propanoate (NaC3H5O2):
Since sodium propanoate (NaC3H5O2) is a salt of a weak acid (propanoic acid), it will undergo hydrolysis in water, leading to the formation of OH- ions. It won't produce H+ ions directly. To calculate the pH, you need to consider the hydrolysis reaction:
C3H5O2- (aq) + H2O (l) β C3H5OH (aq) + OH- (aq)
The reaction results in the production of OH- ions, which affect the pH of the solution. However, the extent of hydrolysis depends on the strength of the conjugate acid (propanoic acid).
(c) Calculation for pure H2O:
Pure water has a neutral pH of 7 because it dissociates to an equal extent into H+ (hydronium) and OH- ions. Since the concentration of both H+ and OH- ions in pure water is equal, the pH is 7.
(d) Calculation for 0.300 M HC3H5O2 and 0.300 M NaC3H5O2:
To calculate the pH of a solution containing both propanoic acid and sodium propanoate, you need to consider the common ion effect. The presence of a common ion (C3H5O2-) in the sodium propanoate solution will suppress the ionization of propanoic acid.
You can use an ICE (Initial, Change, Equilibrium) table to calculate the concentration of H+ ions in the solution. Assume x is the concentration of H+ ions:
Initial Concentration:
[HC3H5O2] = 0.300 M
[NaC3H5O2] = 0.300 M
Change:
[H+] = -x
[C3H5O2-] = +x (from the dissociation of NaC3H5O2)
Equilibrium Concentration:
[HC3H5O2] = 0.300 - x
[C3H5O2-] = x
Use the equation for the dissociation constant (Ka) to find the value of x:
Ka = [H+][C3H5O2-]/[HC3H5O2]
1.3 x 10^-5 = x * x / (0.300 - x)
This is a quadratic equation that can be solved to find the value of x, which represents the concentration of H+ ions. Once you find x, you can calculate the pH using the pH equation: pH = -log[H+].
Note: The concentration values provided are hypothetical and for illustration purposes only. Use the actual concentrations given in the problem to solve the equations.