1.Calculate the PH of 0.025M solution of ammonia kb=1.75x10^-5.
2.The ionization constant of water is Kw=1.00x10^-13.60 at body temperature,37°C .what are the H3O^+ and OH^- concentration at that temperature?
1. To calculate the pH of a solution of ammonia, we first need to find the concentration of hydroxide ions (OH-) in the solution.
Ammonia (NH3) can undergo a reaction with water to produce ammonium (NH4+) and hydroxide (OH-) ions according to the equation: NH3 + H2O -> NH4+ + OH-
Since ammonia is a weak base, we can assume that the concentration of hydroxide ions produced is equal to the concentration of ammonia (NH3) that reacts.
Given that the concentration of ammonia is 0.025M, the concentration of hydroxide ions is also 0.025M.
The hydroxide ion concentration can be used to calculate the pOH of the solution, using the formula: pOH = -log[OH-].
pOH = -log(0.025) = 1.60
The pH of the solution can be calculated using the formula: pH + pOH = 14.
pH + 1.60 = 14
pH = 12.40
Therefore, the pH of the 0.025M solution of ammonia is 12.40.
2. At body temperature, 37°C, the ionization constant of water (Kw) is known to be 1.00x10^-13.6.
The equation for the ionization of water is: H2O <=> H3O+ + OH-
At equilibrium, the concentration of H3O+ equals the concentration of OH-. Let x be the concentration of H3O+ (or OH-).
Using the Kw expression, we can write: Kw = [H3O+][OH-] = 1.00x10^-13.6
Since [H3O+] = [OH-] = x, we can simplify the expression to: x^2 = 1.00x10^-13.6
Taking the square root of both sides, we get: x = (1.00x10^-13.6)^(1/2)
x ≈ 1.00x10^-6.8
Therefore, at a body temperature of 37°C, the concentration of both H3O+ and OH- in water is approximately 1.00x10^-6.8 M.
1. To calculate the pH of a solution of ammonia, we need to first calculate the concentration of hydroxide ions (OH^-) in the solution.
The reaction between ammonia (NH3) and water (H2O) can be represented as follows:
NH3 + H2O ⇌ NH4+ + OH^-
Since the Kb value for ammonia is given as 1.75x10^-5, we can set up an expression for Kb as follows:
Kb = [NH4+][OH^-] / [NH3]
We are given that the concentration of ammonia, [NH3], is 0.025M. Since the initial concentration of NH4+ is 0M, we can assume that x mol/L of NH3 will react with x mol/L of H2O to form x mol/L of NH4+ and x mol/L of OH^-. Using this assumption, we can set up the following equilibrium expression:
Kb = (x)(x) / (0.025 - x)
Since x is assumed to be small compared to 0.025, we can approximate 0.025 - x as 0.025. Simplifying the expression, we get:
1.75x10^-5 = x^2 / 0.025
Rearranging the equation, we have:
x^2 = (1.75x10^-5)(0.025)
x^2 = 4.375x10^-7
Taking the square root of both sides, we get:
x = √(4.375x10^-7)
x = 6.61x10^-4
The concentration of OH^- in the solution is approximately 6.61x10^-4M.
To find the pH, we can use the equation:
pH = -log[H3O+]
Since the solution is basic, we can assume that [H3O+] ≈ Kw / [OH^-]. At 25°C, Kw is equal to 1.00x10^-14. Since the temperature is 37°C, we need to adjust the value of Kw using the temperature coefficient, which is given by the equation:
Kw = Kw25 * 10^(-2pH × ∆T/10)
Substituting the values, we get:
1.00x10^-14 = (1.00x10^-13.60) * 10^(-2pH × (37-25)/10)
Simplifying the equation, we have:
1.00x10^-14 = 10^(-0.136 * pH)
Taking the logarithm of both sides, we get:
-14 = -0.136 * pH
Solving for pH, we have:
pH = -14 / -0.136
pH ≈ 103.03
Therefore, the pH of the solution is approximately 103.03.