a) Calc. PH of 0.010 M HF and 0.030 M KF
e) calc. PH Zn(OH)2 in wAter
a is a buffer. Weak acid and salt of the weak acid. Use the Henderson-Hasselbalch equation.
b. Zn(OH)2 is a Ksp problem. Solve for solubility Zn(OH)2, get the OH from that and convert to pH.
a) To calculate the pH of a solution containing a weak acid (HF) and its conjugate base (KF), we need to consider the dissociation of the weak acid and the common ion effect.
First, let's write down the dissociation equation for HF in water:
HF (aq) ⇌ H+ (aq) + F- (aq)
Since HF is a weak acid, it only partially dissociates in water. Let's assume x is the concentration of H+ ions. Then, the equilibrium concentrations will be:
[H+] = x
[F-] = x
Given that the initial concentration of HF is 0.010 M, the concentration of HF remaining after dissociation will be (0.010 - x) M.
Now, let's consider the common ion effect due to the presence of KF. The F- ions come from the dissociation of KF, so the concentration of F- ions from KF will be 0.030 M.
Using the equilibrium constant expression for the dissociation of HF:
Ka = ([H+][F-]) / [HF]
Substituting the equilibrium concentrations, we have:
Ka = (x * x) / (0.010 - x)
We can assume that x is small compared to 0.010 (the initial concentration of HF), so we can simplify the expression as:
Ka = x^2 / 0.010
Now, we need to solve for x in this equation. Since Ka is an acid dissociation constant, it is usually given. Without the value of Ka, it is not possible to find the exact pH of the solution. However, if you have the Ka value, you can use the quadratic formula to solve for x and then calculate the pH.
e) To calculate the pH of Zn(OH)2 in water, we first need to consider the dissociation of Zn(OH)2.
Zn(OH)2 (s) ⇌ Zn^2+ (aq) + 2OH- (aq)
Zn(OH)2 is a sparingly soluble base, so it only partially dissociates in water. Let's assume x is the concentration of OH- ions. Then, the equilibrium concentrations will be:
[OH-] = x
[Zn^2+] = x
Since Zn(OH)2 is a base, it will react with water to a small extent, producing additional OH- ions. However, in dilute solutions, the concentration of water is relatively constant, so the contribution of further OH- ions is considered negligible.
The concentration of OH- ions can be used to calculate the pOH, and then the pH of the solution. However, to calculate the concentration of OH- ions, we need additional information, such as the solubility product constant (Ksp) for Zn(OH)2. Without this information, it is not possible to determine the exact pH of Zn(OH)2 in water.
If you have the Ksp value for Zn(OH)2, you can use it to calculate the concentration of OH- ions and then convert it to pH.