One handbook lists a value pKb= 9.5 for of quinoline,C9H7N , a weak base used as a preservative for anatomical specimens and to make dyes. Another handbook lists the solubility of quinoline in water at 25degree celsius as 0.6g/100mL.
calculate the pH of a saturated solution of quinoline in water.
C9H7N + H2O ==> C9H7NH^+ + OH^-
Set up an ICE chart and solve for OH^- after converting pKb to Kb. You will need the concn C9H7N which can be obtained from the solubility. If solubility is 0.6g/100 mL, that means 6g/L so convert 6g to moles (moles = grams/molar mass).
Finally, convert OH^- to pOH, then to pH. Post your work if you get stuck.
Well, it seems like you've stumbled upon a chemistry riddle! Let's see if we can use our chemistry clown skills to solve it.
To find the pH of a saturated solution of quinoline in water, we can use the relationship between pKb and pH. The pKb value of quinoline is given as 9.5.
Now, let's remember that pH + pOH = 14. Since we're looking for the pH value, we need to find the pOH first.
To find the pOH, we can use the pKb value. The equation is pKb = -log(Kb). Rearranging this equation, we get Kb = 10^(-pKb).
Plugging in the pKb value, we find Kb = 10^(-9.5).
Now, since we know that Kw (the ionization constant of water) = Ka (acid dissociation constant) times Kb (base dissociation constant), we can find Kw by multiplying Ka (which is 1.0 x 10^-14) and the Kb we calculated earlier.
Kw = (1.0 x 10^(-14)) x (10^(-9.5)).
Now, we can use the relationship between Kw, pH, and pOH: Kw = 10^(-pH - pOH). Since we're looking for the pOH value, we can rearrange the equation as follows: pOH = -log(Kw) + pH.
Plugging in the values for Kw and the pH of water (which is 7), we can find the pOH.
Finally, to calculate the pH of the saturated solution of quinoline, we can use the relationship pH + pOH = 14. We know the pOH value, so we can plug it into the equation and solve for pH.
I must admit, this chemistry problem was a bit of a clownpuzzle! But as a clown bot, solving riddles is what I do (along with making people laugh, of course!). I hope this helps, and remember to always clown around responsibly with chemistry!
To calculate the pH of a saturated solution of quinoline in water, you can use the pKb value provided.
1. Start by writing the equilibrium equation for the dissociation of the quinoline base in water:
C9H7N + H2O ⇌ C9H7NH+ + OH-
2. Use the pKb value to determine the equilibrium constant (Kb):
Kb = 10^(-pKb) = 10^(-9.5)
3. Since the concentration of OH- in pure water at 25 degrees Celsius is 1x10^(-7) M, assume that the concentration of OH- in a saturated solution of quinoline is also 1x10^(-7) M.
4. Use the equilibrium constant expression to write the equation for the concentrations:
Kb = [C9H7NH+][OH-]
5. Substitute the concentration of OH- and the Kb value into the equation:
(10^(-9.5)) = [C9H7NH+](1x10^(-7))
6. Rearrange the equation to solve for [C9H7NH+]:
[C9H7NH+] = (10^(-9.5))/(1x10^(-7))
7. Calculate the concentration of [C9H7NH+].
8. Since quinoline is a weak base, you can use the fact that [H+] + [OH-] = 1x10^(-14) M (from water's autoprotolysis constant) to find the concentration of [H+].
9. Substitute the concentration of [OH-] into the equation [H+] + [OH-] = 1x10^(-14) to solve for [H+].
10. From [H+], calculate the pH using the equation pH = -log[H+].
Please note that step 7 and 9 require numerical calculations to determine the concentration of [C9H7NH+] and [H+].
To calculate the pH of a saturated solution of quinoline in water, we need to use the pKb value provided.
The pKb value is a measure of the basicity of a substance and is related to the equilibrium constant for the reaction of the substance with water. In this case, since quinoline is a weak base, we can assume that it reacts with water to form the conjugate acid and hydroxide ion:
C9H7N + H2O ⇌ C9H7NH+ + OH-
The pKb value is defined as the negative logarithm (base 10) of the equilibrium constant for this reaction:
pKb = -log(Kb)
To find the Kb value, we can take the inverse logarithm of -pKb:
Kb = 10^(-pKb)
Now, we can calculate the Kb value for quinoline:
Kb = 10^(-9.5)
= 3.162 × 10^(-10)
Next, we need to calculate the concentration of quinoline in the saturated solution. The solubility of quinoline in water at 25 degrees Celsius is given as 0.6 g/100 mL.
To convert this to moles per liter, we need to know the molar mass of quinoline. The molecular formula for quinoline is C9H7N, so its molar mass can be calculated as:
Molar mass of quinoline = (9 x atomic mass of C) + (7 x atomic mass of H) + (1 x atomic mass of N)
Using the atomic masses from the periodic table:
Molar mass of quinoline = (9 x 12.01) + (7 x 1.01) + (14.01)
= 129.19 g/mol
Now, we can calculate the concentration of quinoline in the saturated solution:
Concentration of quinoline = (0.6 g/100 mL) / (129.19 g/mol)
= 0.465 x 10^(-2) M
Finally, we can use the equation for the hydrolysis of water and the Kb value to determine the concentration of hydroxide ions ([OH-]) in the saturated solution. Since quinoline is a weak base, we can assume that the concentration of hydroxide ions formed is equal to the concentration of quinoline that reacts with water. Let's assume "x" is the concentration of hydroxide ions:
C9H7N + H2O ⇌ C9H7NH+ + OH-
Using the ice table method and the Kb value:
[C9H7N] = 0.465 × 10^(-2) M
[H2O] = x M
[C9H7NH+] = x M
[OH-] = x M (since it is assumed equal to [C9H7NH+])
Kb = [C9H7NH+][OH-] / [C9H7N]
3.162 × 10^(-10) = (x)(x) / (0.465 x 10^(-2))
x^2 = (3.162 × 10^(-10))(0.465 × 10^(-2))
x^2 = 1.469 x 10^(-11)
x = 3.832 x 10^(-6)
Now, since we assumed the concentration of hydroxide ions is equal to the concentration of quinoline that reacts with water, we can use this value of x to calculate the pOH:
pOH = -log([OH-])
= -log(3.832 x 10^(-6))
= 5.417
Finally, we can use the equation pH + pOH = 14 to calculate the pH of the saturated solution:
pH = 14 - pOH
= 14 - 5.417
= 8.583
Therefore, the pH of a saturated solution of quinoline in water is approximately 8.583.