The basicity constant of pyridine can be expressed as pKb = 8.75 what will be the pH of a 1M (1 mole per liter) solution of pyridine in water?
Py + HOH ==> PyH^+ + OH^-
Kb = (PyH^+)(OH^-)/(Py)
Change pKb to Kb, plug in x for (PyH^+) and (OH^-) and 1 for (Py) and solve for x. That will be OH^- so determine pOH and remember pH + pOH = 14.
Post your work if you get stuck.
The pKb of pyridine is 8.75. What is the pH of a 0.480M solution of pyridine?
To find the pH of a 1M solution of pyridine in water, we need to determine the concentration of hydroxide ions (OH-) in the solution first.
Given that the basicity constant (Kb) of pyridine is expressed as pKb = 8.75, we can convert it to Kb using the equation:
Kb = 10^(-pKb)
Kb = 10^(-8.75)
Kb ≈ 1.77 x 10^(-9)
Now, let's assume the concentration of pyridine (Py) is 1M, the concentration of hydroxide ions (OH-) is x, and the concentration of pyridinium ions (PyH+) is also x.
The dissociation reaction of pyridine in water is as follows:
Py + HOH → PyH+ + OH-
The Kb expression is given by:
Kb = (PyH+)(OH-) / (Py)
Substituting the known values, we have:
1.77 x 10^(-9) = (x)(x) / 1
Simplifying the equation:
1.77 x 10^(-9) = x^2
Taking the square root of both sides:
x ≈ 1.33 x 10^(-5)
Therefore, the concentration of hydroxide ions (OH-) in the solution is approximately 1.33 x 10^(-5) M.
To determine the pOH, we use the following relationship:
pOH = -log(OH-)
pOH = -log(1.33 x 10^(-5))
pOH ≈ 4.88
Finally, since pH + pOH = 14, we can find the pH:
pH = 14 - pOH
pH ≈ 14 - 4.88
pH ≈ 9.12
Therefore, the pH of a 1M solution of pyridine in water is approximately 9.12.
To find the pH of a 1M solution of pyridine in water, we first need to convert the pKb value to Kb. The formula to convert pKb to Kb is Kb = 10^(-pKb).
In this case, pKb = 8.75, so we can calculate Kb as follows:
Kb = 10^(-8.75)
Next, we plug in x for the concentrations of PyH+ and OH-, and 1 for the concentration of Py (which is 1M in a 1M solution). Then we can set up the equation for Kb as:
Kb = (x)(x)/(1)
Simplifying this equation gives us:
Kb = x^2
Now we can solve for x by taking the square root of both sides:
sqrt(Kb) = x
To find the concentration of OH-, we need to subtract the concentration of x from the original concentration of pyridine. So the concentration of OH- is (1M - x).
To find the pOH, we need to take the negative logarithm (base 10) of the concentration of OH-. We can use the formula pOH = -log10([OH-]).
Finally, we can calculate the pH using the equation pH + pOH = 14. Subtract the pOH value we found from 14 to obtain the pH.
Remember to post your work if you get stuck, and let me know if you have any further questions!