Determine the pH of each of the following solutions.
0.21 M KCHO2
0.19 M CH3NH3I
0.19 M KI
I'm not sure how i should set up the ice chart, and what Ka I should be using for the solutions
To determine the pH of each solution, we need to consider the dissociation of the weak acids or bases present in the solutions. In this case, we will need to use the appropriate equilibrium constant (Ka) for each solution.
1. 0.21 M KCHO2:
KCHO2 is the salt of a weak acid (HCHO2) and a strong base (KOH). When KCHO2 dissolves in water, it undergoes hydrolysis. The hydrolysis reaction is as follows:
KCHO2 (aq) + H2O (l) ⇌ K+ (aq) + CHO2- (aq) + H3O+ (aq)
Since CHO2- is the conjugate base of the weak acid HCHO2, we will use the equilibrium constant for the acid dissociation, Ka.
2. 0.19 M CH3NH3I:
CH3NH3I is a salt of a weak base (CH3NH2) and a strong acid (HI). When CH3NH3I dissolves in water, it undergoes hydrolysis. The hydrolysis reaction is as follows:
CH3NH3I (aq) + H2O (l) ⇌ CH3NH3+ (aq) + I- (aq) + H3O+ (aq)
Since CH3NH3+ is the conjugate acid of the weak base CH3NH2, we will use the equilibrium constant for the base dissociation, Kb.
3. 0.19 M KI:
KI is the salt of a strong acid (HI) and a strong base (KOH). When KI dissolves in water, it completely dissociates into K+ and I- ions. Since both ions do not further participate in any acid-base reactions, there is no need to use an equilibrium constant here.
Now, let's determine the pH for each solution step by step:
1. For 0.21 M KCHO2:
- Set up the ice chart, considering the hydrolysis reaction:
KCHO2 (aq) + H2O (l) ⇌ K+ (aq) + CHO2- (aq) + H3O+ (aq)
- Initially, you have 0.21 M of KCHO2 and 0 M of H3O+ and CHO2-.
- At equilibrium, let's assume the concentration of H3O+ is x M.
- This means the concentration of CHO2- will also be x M.
- The concentration of K+ remains 0.21 M since it does not participate in the reaction.
- The KCHO2 concentration will be 0.21 - x M.
- Now, write the expression for Ka: Ka = ([H3O+] [CHO2-]) / [KCHO2].
- Substitute the given values and solve for x.
- Finally, calculate the pH using the concentration of H3O+.
2. For 0.19 M CH3NH3I:
- Set up the ice chart, considering the hydrolysis reaction:
CH3NH3I (aq) + H2O (l) ⇌ CH3NH3+ (aq) + I- (aq) + H3O+ (aq)
- Initially, you have 0.19 M of CH3NH3I and 0 M of H3O+ and CH3NH3+.
- At equilibrium, let's assume the concentration of H3O+ is x M.
- This means the concentration of CH3NH3+ will also be x M.
- The concentration of I- remains 0.19 M since it does not participate in the reaction.
- The CH3NH3I concentration will be 0.19 - x M.
- Now, write the expression for Kb: Kb = ([CH3NH3+] [H3O+]) / [CH3NH3I].
- Substitute the given values and solve for x.
- Finally, calculate the pH using the concentration of H3O+.
3. For 0.19 M KI:
- Since KI is the salt of a strong acid and a strong base, it completely dissociates into K+ and I-.
- The concentration of both K+ and I- would be 0.19 M.
- However, neither ion undergoes any acid-base reactions with water, so there is no need to determine the pH.
Remember to use the appropriate equilibrium constant for each reaction and perform the necessary calculations to determine the pH for each solution.
Why did the banana go to the doctor? Because it wasn't peeling well!
To determine the pH of each solution, we need to consider the acid-base properties of the solutes.
Let's start with 0.21 M KCHO2. This compound is the salt formed by the reaction between potassium hydroxide (KOH) and formic acid (HCHO2). Formic acid is a weak acid, and its dissociation can be represented as follows:
HCHO2 ⇌ H+ + CHO2-
The potassium cation (K+) is a spectator ion and does not affect the pH. Thus, we are left with the conjugate base of formic acid, CHO2-, which can be considered a weak base. The equilibrium equation for the hydrolysis of CHO2- is:
CHO2- + H2O ⇌ HCHO2 + OH-
Since we have the concentration of the CHO2- ion (0.21 M), we can set up an ICE chart:
Initial: CHO2- = 0.21 M, H2O = 0 M, OH- = 0 M
Change: -x, +x, +x
Equilibrium: 0.21 - x, x, x
Now, we can set up the expression for the base ionization constant (Kb):
Kb = (HCHO2)(OH-) / (CHO2-)
At equilibrium, the concentrations of HCHO2 and OH- are both x. Substituting the equilibrium concentrations into the Kb expression:
Kb = x * x / (0.21 - x)
Since Kb is a small value, we can assume that x is much smaller compared to 0.21, so we can approximate (0.21 - x) to approximately 0.21:
Kb ≈ x * x / 0.21
Now, we can use Kb to calculate the concentration of OH-, and from there, determine the pOH and pH of the solution. However, since formic acid is a weak acid, Kb is very small, and the concentration of OH- will be even smaller. So, in this case, we can assume that [OH-] ≈ 0, and therefore, the solution will be slightly acidic.
Now, for the other two solutions (0.19 M CH3NH3I and 0.19 M KI), they don't have any known acidic or basic properties. So, we cannot determine their pH without more information.
Why did the scarecrow win an award? Because he was outstanding in his field!
To determine the pH of each solution, we need to understand the properties of the solutes and their acid-base behavior. Let's break down each solution and assess how to set up the ICE chart and which Ka value to use.
1. 0.21 M KCHO2:
KCHO2 dissociates to form the acetate ion (CH3COO-) in solution. Acetate is the conjugate base of acetic acid, so we will be using its acid dissociation constant, Ka.
- Set up the ICE chart:
KCHO2 dissociates in water according to the equation: KCHO2 ⇌ K+ + CHO2-
The initial concentrations are:
[KCHO2] = 0.21 M
[K+] = 0 M (as it is a spectator ion)
[CHO2-] = 0 M (since it is not present initially)
The change in concentration will be -x for both [KCHO2] and [CHO2-] and +x for [K+], as 1 mole of KCHO2 dissociates to form 1 mole of K+ and CHO2-.
The equilibrium concentrations will be:
[K+] = x
[CHO2-] = x
[KCHO2] = 0.21 M - x
Considering the expression of Ka for acetic acid:
Ka = [CH3COO-][H3O+]/[CH3COOH]
We will neglect the contribution of water, assuming it remains constant.
2. 0.19 M CH3NH3I:
CH3NH3I dissociates into methylamine, CH3NH2, and hydriodic acid, HI. We need to consider the dissociation of methylamine, which is a weak base.
- Set up the ICE chart:
CH3NH3I ⇌ CH3NH3+ + I-
The initial concentrations are:
[CH3NH3I] = 0.19 M
[CH3NH3+] = 0 M (initially not present)
[I-] = 0 M (initially not present)
The change in concentration will be -x for [CH3NH3I] and +x for both [CH3NH3+] and [I-] since 1 mole of CH3NH3I dissociates to form 1 mole of CH3NH3+ and I-.
The equilibrium concentrations will be:
[CH3NH3+] = x
[I-] = x
[CH3NH3I] = 0.19 M - x
Considering the expression of Kb for methylamine:
Kb = [CH3NH3+][OH-]/[CH3NH2]
Since we are looking for the pH, which is related to the concentration of H3O+ (or [H+]), we can use the expression:
Kw = [H+][OH-] = 1.0 x 10^-14
Using the relation: Kw = Ka x Kb, we can find [H+].
3. 0.19 M KI:
KI dissociates to form the iodide ion (I-) in solution. Iodide ion is the conjugate base of hydriodic acid (HI), which is a strong acid. Since HI is already fully dissociated, we do not need to set up an ICE chart for this solution.
To determine the pH of a strong acid solution like KI, we can directly assume that all of the I- will become [H+], resulting in a pH equal to -log[H+] concentration.
Please note: Before calculating the pH, it's important to determine the degree of dissociation (x) for each case by assessing the strength of the acid or base involved. The Ka and Kb expressions given above can help you find x, which will then allow you to calculate the equilibrium concentrations and ultimately the pH.
Remember, the calculation steps may involve solving quadratic equations in some cases where x^2 terms appear.
For the first one.
This is potassium formate, a salt. The K ion is not hydrolyzed but the formate ion is in which it acts as a base.
CHO2^- + HOH ---> HCHO2 + OH^-
Kb = (Kw/Ka) = (OH^-)(HCHO2)/(CHO2^-)
So HCHO2 is x, OH^- is x, CHO2^- is 0.21-x, Kw you know and Ka is Ka for HCHO2 (formic acid). Solve for x which will be OH^-, convert to pOH then to pH.
The second one is a salt, also, and it is the cation that hydrolyzes as follows:
CH3NH3^+ + H2O ==> CH2NH2 + H3O^+ but it is worked the same way as the formate salt.
For the last one you must recognize that neither the K ion nor the I ion is ionized; therefore, the solution is just like table salt in water and the pH is determined by the ionization of water. pH = 7.0