Calculate the mass of HONH2 required to dissolve in enough water to make 255.5 mL of solution having a pH of 10.02 (Kb = 1.110−8).
Well, let's calculate that, but first let me grab my calculator... or maybe I should just juggle some numbers instead! 🤡
To find the mass of HONH2 required, we need to use the formula for the hydroxide ion concentration in a basic solution, which is given by:
OH- = sqrt(Kb * [HONH2])
Where Kb is the base dissociation constant (1.110^-8) and [HONH2] is the concentration of HONH2. Since we want to make a solution with pH 10.02, we know that [H+] (hydrogen ion concentration) will be 10^(-10.02).
To find [OH-], we can use the fact that in water, [H+] * [OH-] = Kw, which is the ion product of water (1.0 x 10^(-14)).
Using the values we have, [OH-] = Kw / [H+] = (1.0 x 10^(-14)) / (10^(-10.02))
Now that we have [OH-], we can substitute into the first equation to solve for [HONH2]:
sqrt(Kb * [HONH2]) = [OH-]
squaring both sides:
Kb * [HONH2] = [OH-]^2
Let's calculate [HONH2]:
[HONH2] = ([OH-]^2) / Kb
And the mass of HONH2 can be calculated using the formula:
mass = (volume * concentration * molar mass) / 1000
I know that was a lot of juggling with numbers and formulas, but don't worry, I've got it all under control! 🤹♀️
Just let me calculate those values and I'll give you the mass of HONH2 required.
To calculate the mass of HONH2 required, we first need to understand the relationship between pH and the concentration of the base in a solution.
Step 1: Calculate the pOH of the solution.
The pOH is related to the pH by the equation: pOH = 14 - pH
In this case, the given pH is 10.02, so the pOH is: pOH = 14 - 10.02 = 3.98
Step 2: Calculate the concentration of OH- ions in the solution.
The concentration of OH- can be found using the equation: OH- concentration = 10^(-pOH)
In this case, the concentration of OH- is: OH- concentration = 10^(-3.98)
Step 3: Calculate the concentration of the base (HONH2) in the solution.
Since HONH2 is a weak base, we can use the Kb value to find the concentration of the base. Kb is the equilibrium constant for the base, and it is defined as the ratio of the concentration of the products (NH2- and H2O) to the concentration of the reactant (HONH2) in the dissociation reaction.
The equation for the dissociation of HONH2 is: HONH2 + H2O ↔ NH2- + H3O+
The Kb expression for this reaction is: Kb = [NH2-][H3O+] / [HONH2]
Given the Kb value (Kb = 1.110^-8), we can rearrange the equation to solve for the concentration of HONH2.
Step 4: Calculate the concentration of HONH2 in the solution using the Kb value.
[HONH2] = [NH2-][H3O+] / Kb
Since we know the concentration of OH- ions, we also know the concentration of H3O+ ions in the solution because of the autoionization of water (H2O ↔ H+ + OH-). Therefore, [H3O+] = [OH-].
[HONH2] = [NH2-][OH-] / Kb
Step 5: Calculate the moles of HONH2 in the solution.
The concentration of HONH2 is given in moles per liter (M), so we can convert it to moles using the equation:
moles of HONH2 = [HONH2] * volume of the solution in liters
In this case, the volume of the solution is given as 255.5 mL, so we need to convert it to liters by dividing by 1000: volume of the solution = 255.5 mL / 1000 = 0.2555 L
moles of HONH2 = [HONH2] * volume of the solution in liters
Step 6: Calculate the mass of HONH2 required.
The molar mass of HONH2 can be found by adding up the molar masses of each element in the compound: H = 1 g/mol, O = 16 g/mol, N = 14 g/mol
Molar mass of HONH2 = 1(1) + 1(16) + 1(14) + 1(1) = 32 g/mol
mass of HONH2 = moles of HONH2 * molar mass of HONH2
Now we can substitute the values into the equation to calculate the mass of HONH2 required.
To calculate the mass of HONH2 required to dissolve in water, we need to find the number of moles of HONH2 needed first. Then, we can convert the moles to grams using the molar mass of HONH2.
Here's a step-by-step solution:
Step 1: Calculate the hydroxide ion concentration (OH-) using the pH.
pOH = 14 - pH
pOH = 14 - 10.02
pOH = 3.98
[OH-] = 10^(-pOH)
[OH-] = 10^(-3.98)
[OH-] = 1.26 x 10^(-4) M
Step 2: Calculate the concentration of HONH, assuming it is completely ionized.
[OH-] = [HONH2]
[HONH2] = 1.26 x 10^(-4) M
Step 3: Calculate the number of moles of HONH2 required.
moles = concentration x volume (in liters)
moles = 1.26 x 10^(-4) M x 0.2555 L
moles = 3.2223 x 10^(-5) mol
Step 4: Find the molar mass of HONH2.
H: 1 mol x 1g/mol = 1 g
O: 1 mol x 16 g/mol = 16 g
N: 1 mol x 14 g/mol = 14 g
H2: 2 mol x 1 g/mol = 2 g
Molar mass of HONH2 = 1g + 16g + 14g + 2g = 33 g/mol
Step 5: Convert the moles to grams.
mass = moles x molar mass
mass = 3.2223 x 10^(-5) mol x 33 g/mol
mass = 0.001063 g
Therefore, the mass of HONH2 required to dissolve in enough water to make a 255.5 mL solution with a pH of 10.02 is approximately 0.001063 grams.
This is what I want to do but I am not sure.
HONH2 + H2O---> OH + H2ONH2
14-pH=pOH=3.98
10^(-3.98)=OH concentration=1.05 x 10^-4 M
1.1 x 10^-8=Kb=[1.05 x 10^-4 M][1.05 x 10^-4 M]/[x-1.05 x 10^-4 M]
Solve for x, 5% rule allows us to eliminate 1.05 x 10^-4 M from the denominator.
([1.05 x 10^-4 M][1.05 x 10^-4 M]/1.1 x 10^-8)=x
x =molarity of HONH2
Molarity of HONH2*(255.5 x 10^-3L)= moles of HONH2
moles of HONH2*(47.013 g of HONH2/mol)= mass of HONH2