a)Calculate the pH of a solution of 0.75 M HNO3, which is a strong acid.
b)Calculate the pH of a 1.0 L aqueous solution made from 0.80 mol CH3COOH and 0.20 mol
KCH3COO (Ka = 1.8 x 10^-5).
a) To calculate the pH of a solution of a strong acid like HNO3, you need to use the formula for pH, which is given by:
pH = -log[H+]
In this case, HNO3 is a strong acid, which means it dissociates completely in water to produce H+ ions. So, the concentration of H+ ions is equal to the concentration of the HNO3 solution.
Given that the solution is 0.75 M HNO3, the concentration of H+ ions is 0.75 M. Plugging this value into the pH formula, we get:
pH = -log(0.75)
Using the logarithm function on a calculator or computer software, you can find that the value of -log(0.75) is approximately 0.12.
Therefore, the pH of the given 0.75 M HNO3 solution is approximately 0.12.
b) To calculate the pH of a solution made from a weak acid and its conjugate base, you need to consider the acid dissociation constant, Ka, for the weak acid.
In this case, you have a solution made from CH3COOH (acetic acid) and KCH3COO (potassium acetate). The equilibrium reaction for the dissociation of acetic acid is:
CH3COOH ⇌ CH3COO- + H+
The Ka value for acetic acid is given as 1.8 x 10^-5.
To solve this problem, you need to set up an ICE (Initial, Change, Equilibrium) table and use the equilibrium expression for Ka:
Ka = [CH3COO-][H+]/[CH3COOH]
Given that you have 0.80 mol CH3COOH and 0.20 mol KCH3COO in 1.0 L of solution, the initial concentrations of CH3COO- and H+ are both 0 mol, and the initial concentration of CH3COOH is 0.80 M.
In the equilibrium state, the change in concentration of CH3COOH is -x, and the change in concentration of CH3COO- and H+ is +x (since they are formed as the acid dissociates).
Setting up the ICE table:
CH3COOH + H2O ⇌ CH3COO- + H+
Initial: 0.80 M 0 M 0 M
Change: -x -x +x
Equilibrium:0.80 - x x x
Now, substitute the equilibrium concentrations into the equilibrium expression for Ka:
Ka = [CH3COO-][H+]/[CH3COOH]
1.8 x 10^-5 = (x)(x)/(0.80 - x)
Since it is a weak acid, we can assume that x is very small compared to 0.80. Therefore, we can approximate 0.80 - x as 0.80.
1.8 x 10^-5 = x^2/0.80
Rearranging the equation, we get:
x^2 = 0.80 * 1.8 x 10^-5
x = √(0.80 * 1.8 x 10^-5)
Using a calculator or computer software, you can find the value of x, which is approximately 0.0085 M.
Now, to find the pH, you need to calculate the concentration of H+ ions. Since the concentration of H+ ions is equal to x, the pH can be calculated using the formula:
pH = -log[H+]
Plugging in the value of x, we get:
pH = -log(0.0085)
Calculating this using a calculator or computer software, you will find that the pH of the given solution is approximately 2.07.