1. What is the [K+] of a solution prepared by dissolving 0.140 g of potassium hydroxide in
sufficient pure water to prepare 250.0 ml of solution?
2. What is the pH of a 0.570 M solution of aniline? Kb = 7.4 × 10-10
Convert 0.140 g KOH to mols. mols = grams/molar mass.
There is 1 mol K^+ per mol KOH.
Then [K^+] = mols/0.250 L = ??
Let's designate, for simplicity in typing, aniline as RNH2.
RNH2 + HOH ==> RNH3^+ + OH^-
Kb = (RNH3^+)(OH^-)/(RNH2) = 7.4 x 10^-10
(RNH3^+) = x
(OH^-) = x
(RNH2) = 0.570 - x
Substitute into Kb and solve for x = (OH^-), convert to (H^+) and solve for pH. Post your work if you get stuck.
Check my thinking.
1. To find the [K+] (concentration of potassium ions) in the solution, we can use the formula for molarity.
First, we need to calculate the number of moles of potassium hydroxide (KOH) in 0.140 g. We can use the molar mass of KOH to convert grams to moles.
The molar mass of KOH is:
K: 39.10 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
So, the molar mass of KOH is 39.10 + 16.00 + 1.01 = 56.11 g/mol.
Now, we can calculate the number of moles (n) using the formula:
n = mass / molar mass
n = 0.140 g / 56.11 g/mol = 0.00249 mol
Next, we need to calculate the molarity (M) of the solution using the formula:
M = moles / volume
M = 0.00249 mol / 0.250 L = 0.00996 M
Therefore, the concentration of potassium ions [K+] in the solution is 0.00996 M.
2. To calculate the pH of the solution of aniline, we need to use the equilibrium constant (Kb) for the reaction between aniline and water, which gives the concentration of hydroxide ions (OH-) in the solution.
The equation for the reaction between aniline (C6H5NH2) and water (H2O) is:
C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-
The Kb value for aniline is given as 7.4 × 10^-10.
To find the concentration of OH-, we can first find the concentration of aniline (C6H5NH2). Since the solution's concentration is given as 0.570 M, we know that the concentration of aniline is also 0.570 M.
Using the equation for Kb:
Kb = [C6H5NH3+][OH-] / [C6H5NH2]
We can rearrange the equation to solve for [OH-]:
[OH-] = (Kb * [C6H5NH2]) / [C6H5NH3+]
Substituting the values:
[OH-] = (7.4 × 10^-10 * 0.570 M) / 0.570 M
[OH-] = 7.4 × 10^-10 M
Since the solution contains equal concentrations of [H+] and [OH-] ions due to water's self-ionization, we know that pOH = -log[OH-].
pOH = -log(7.4 × 10^-10) ≈ 9.13
Now, we can find the pH using the formula:
pH = 14 - pOH
pH = 14 - 9.13 ≈ 4.87
Therefore, the pH of a 0.570 M solution of aniline is approximately 4.87.
To find the concentration of potassium ions ([K+]) in the solution, we need to calculate the number of moles of potassium hydroxide (KOH) dissolved in 250.0 ml of solution.
Step 1: Calculate the moles of KOH
Given the mass of potassium hydroxide (KOH) is 0.140 g and the molar mass of KOH is 56.11 g/mol (from the periodic table), we can calculate the moles of KOH using the formula:
moles = mass / molar mass
moles = 0.140 g / 56.11 g/mol
Step 2: Calculate the concentration ([K+])
The concentration is given in moles per liter (mol/L). Since we are given the volume as 250.0 ml, we need to convert it to liters by dividing by 1000:
volume = 250.0 ml / 1000
volume = 0.250 L
Then, we can calculate the concentration ([K+]) using the formula:
concentration = moles / volume
Now we can substitute the values we found into the formula:
concentration = (0.140 g / 56.11 g/mol) / 0.250 L
Finally, calculate the value to find the [K+] concentration.
For the second question, to find the pH of the solution, we need to determine the pKa value for aniline using the given Kb value. The pKa is related to Kb by the formula:
pKa + pKb = 14
Rearranging the formula, we find:
pKa = 14 - pKb
From the given value, Kb = 7.4 × 10-10, we can calculate pKa:
pKa = 14 - log(Kb) = 14 - log(7.4 × 10-10)
Once we have the value of pKa, we can calculate the concentration of the anilinium ion, [C6H5NH3+], using the Henderson-Hasselbalch equation:
pH = pKa + log([C6H5NH3+]/[C6H5NH2])
Since we have the concentration of aniline as 0.570 M, we can calculate the pH using the formula:
pH = pKa + log([C6H5NH3+]/0.570)