Respuesta :
This is an incomplete question, here is a complete question.
A chemistry graduate student is given 500 mL of a 0.20 M pyridine C₅H₅N solution. Pyridine is a weak base with Kb = 1.7 × 10⁻⁹. What mass of C₅H₅NHCl should the student dissolve in the C₅H₅N solution to turn it into a buffer with pH = 4.76 ?
You may assume that the volume of the solution doesn't change when the C₅H₅NHCl is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.
Answer : The mass of C₅H₅NHCl is, 34 grams.
Explanation :
First we have to calculate the value of [tex]pK_b[/tex].
The expression used for the calculation of [tex]pK_b[/tex] is,
[tex]pK_b=-\log (K_b)[/tex]
Now put the value of [tex]K_b[/tex] in this expression, we get:
[tex]pK_b=-\log (1.7\times 10^{-9})[/tex]
[tex]pK_b=8.77[/tex]
Now we have to calculate the value of [tex]pK_a[/tex]
[tex]pK_a+pK_b=pK_w\\\\pK_a+8.77=14\\\\pK_a=5.23[/tex]
Now we have to calculate the concentration of C₅H₅NHCl.
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[C_5H_5N]}{[C_5H_5NHCl]}[/tex]
Now put all the given values in this expression, we get:
[tex]4.76=5.23+\log (\frac{0.20}{[C_5H_5NHCl]})[/tex]
[tex][C_5H_5NHCl]=0.59M[/tex]
Now we have to calculate the mass of C₅H₅NHCl.
[tex]\text{Concentration of }C_5H_5NHCl=\frac{\text{Mass of }C_5H_5NHCl\times 1000}{\text{Molar mass of }C_5H_5NHCl\times \text{Volume of solution (in mL)}}[/tex]
Now put all the given values in this formula, we get:
[tex]0.59=\frac{\text{Molar mass of }C_5H_5NHCl\times 1000}{115.5g/mole\times 500mL}[/tex]
[tex]\text{Molar mass of }C_5H_5NHCl=34.0725g\approx 34g[/tex]
Thus, the mass of C₅H₅NHCl is, 34 grams.
