When 23.5 mL of 0.12 M HCl are added to 50.0 mL of a 0.15 M solution of a weak monoprotic base, the pH of the solution is 10.00. What is the Kb of the weak base?

Question

When 23.5 mL of 0.12 M HCl are added to 50.0 mL of a 0.15 M solution of a weak monoprotic base, the pH of the solution is 10.00. What is the Kb of the weak base?

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3 answers
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Answer 1

DrBob222 answered
5 years ago

millimols HCl = mL x M = 23.5 x 0.12 = ?= approx 3

millimols BOH = 50 x 0.15 = ? appox 7.5
You need to use better accuracy..These are only approx.
............HCl + BOH ==> BCl + H2O
I............3.........7.5...........0..........0
C,,,,,,,,,-3.........-3.............+3.......+3
E...........0.........4.2...........3

You see you have a buffered solution consiting of a weak base and it's acid (BCl). Plug this into the Henderson-Hasselbalch equation, solve for pKa which is the only unknown. Then to get pKb use
pKa + pKb = pKw = 14. You know pKw and pKa, solve for pKb.
Post your work if you get stuck.

Answer 2

Doc48 answered
5 years ago

23.5-ml(0.12 HCl) + 50-ml(0.15M BH:OH)

=> 0.0235(0.12) mole HCl + 0.050(0.15) mole BH:OH
=> 0.00282 mole HCl + 0.0075 mole BH:OH
=> (0.0075 – 0.00282) mole BH:OH + 0.00282 mole BH⁺:Clˉ
=> 0.00468 mole BH:OH + 0.00282 mole BH⁺:Clˉ
=> (0.00468 mole BH:OH/0.0735-L Soln) + (0.00282 mole BH⁺:Clˉ/0.0735-L Soln)
=> (0.0367M) BH:OH + (0.0384M BH⁺:Clˉ) => pH = 10 => pOH = 14 – 10 = 4 => [OHˉ] = 10ˉ⁴M

……… BH:OH ==> BH⁺ + OHˉ
I ….. 0.0367M …….. 0.0384M ……….0
C …….. -X ………………….. +X ……………+X
E …..0.0367-X ……….0.0384+X……….10ˉ⁴M
……=0.0366M ………=0.0385M

K(b) = [BH⁺][:OHˉ]/[BH:OH] = (0.0366)(0.0385)/(10ˉ⁴) = 1.05 x 10ˉ⁴

Answer 3

Explain Bot answered
11 months ago

To find the Kb of the weak base, we need to use the equation for the reaction between the weak base and water:

Base + H2O ⇌ OH- + Acid

First, let's determine the number of moles of HCl that were added to the solution. To do this, we'll use the formula:

moles = concentration (M) * volume (L)

The concentration of the HCl is given as 0.12 M, and the volume is 23.5 mL, which can be converted to liters by dividing by 1000:

moles HCl = 0.12 M * (23.5 mL / 1000) = 0.00282 moles

Next, we'll determine the number of moles of the weak base in the solution. We'll use the formula:

moles = concentration (M) * volume (L)

The concentration of the weak base is given as 0.15 M, and the volume is 50.0 mL, which can be converted to liters by dividing by 1000:

moles weak base = 0.15 M * (50.0 mL / 1000) = 0.00750 moles

Since the reaction is in a 1:1 ratio, the number of moles of OH- formed is equal to the moles of HCl added:

moles OH- = 0.00282 moles

Now, we need to find the concentration of OH- in the solution. We'll divide the moles of OH- by the final volume of the solution:

Concentration OH- = moles OH- / total volume (L)

The total volume is the sum of the volumes of the HCl and weak base solutions, which is 23.5 mL + 50.0 mL, converted to liters:

total volume = (23.5 mL + 50.0 mL) / 1000 = 0.0735 L

Concentration OH- = 0.00282 moles / 0.0735 L = 0.0384 M

Since the pH of the solution is 10.00, we can find the concentration of H+ using the equation:

[H+] = 10^(-pH)

[H+] = 10^(-10.00) = 1.0 x 10^(-10) M

Since the solution is in a 1:1 ratio, the concentration of the weak base, B, is equal to the OH- concentration:

[B] = [OH-] = 0.0384 M

Now, we can use the equation for Kb (the equilibrium constant for the reaction between the weak base and water):

Kb = [OH-] * [H+] / [B]

Kb = (0.0384 M) * (1.0 x 10^(-10) M) / (0.0384 M)

Kb = 1.0 x 10^(-10) ******/

Therefore, the Kb of the weak base is 1.0 x 10^(-10).