Respuesta :
Answer : The pH of the solution is, 9.4
Solution : Given,
Concentration (c) = 0.025 M
Base dissociation constant = [tex]pK_b=8.75[/tex]
First we have to calculate the value of [tex]K_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]pK_b[/tex] in this expression, we get:
[tex]8.75=-\log (K_b)[/tex]
[tex]K_b=1.78\times 10^{-9}[/tex]
The given equilibrium reaction is,
[tex]C_5H_5N+H_2O\rightleftharpoons C_5H_5NH^++OH^-[/tex]
initially conc. 0.335 0 0
At eqm. (0.335-x) x x
Formula used :
[tex]k_b=\frac{[C_5H_5NH^+][OH^-]}{[C_5H_5N]}[/tex]
Now put all the given values in this formula ,we get:
[tex]1.78\times 10^{-9}=\frac{(x)(x)}{(0.335-x)}[/tex]
By solving the terms, we get:
[tex]x=2.4\timees 10^{-5}[/tex]
Now we have to calculate the concentration of hydroxide ion.
[tex][OH^-]=x=2.4\times 10^{-5}M[/tex]
Now we have to calculate the pOH.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (2.4\times 10^{-5})[/tex]
[tex]pOH=4.6[/tex]
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.6\\\\pH=9.4[/tex]
Therefore, the pH of the solution is, 9.4
Answer:
- The pH of the solution of pyridine is [tex]9.28[/tex]
Explanation:
Given:
The [tex]pKb[/tex]of pyridine is [tex]8.75[/tex]
Step 1 of 3
To find the value of [tex]Kb[/tex]
[tex]$-\log \mathrm{Kb}=8.75$[/tex]
[tex]$K_{b}=1.78 \times 10^{-9}$[/tex]
The given equilibrium reaction is,
[tex]$C_{5} H_{5} N+H_{2} O \rightleftharpoons C_{5} H_{5} N H^{+}+O H^{-}$[/tex]
[tex]I ........0.2.......0.........0\\C........x.....+x.......+x\\E.......0.2-x......+x.......+x[/tex]
Step 2 of 3
Now put all the given values in this formula, we get:
[tex]$1.78 \times 10^{-9}=\frac{(x)(x)}{(0.2-x)}$[/tex]
By solving, we get:
[tex]x=1.89\times 10^{-5}$[/tex]
Now we have to calculate the concentration of hydroxide ion.
[tex]$\left[O H^{-}\right]=x=1.89 \times 10^{-5} M$[/tex]
Step 3 of 3
Calculate the [tex]$p O H$[/tex]
[tex]$p O H=-\log \left[O H^{-}\right]$[/tex]
[tex]$p O H=-\log \left(1.89 \times 10^{-5}\right)$\\[/tex]
[tex]$p O H=4.72$[/tex]
Calculate the [tex]$\mathrm{pH}$[/tex] value
[tex]$p H+p O H=14$[/tex]
[tex]$p H=14-p O H$[/tex]
[tex]$p H=14-4.72$[/tex]
[tex]$p H=9.28$[/tex]
Therefore, the pH of the solution of pyridine is [tex]9.28[/tex]
Learn more about Pyridine, refer:
- https://brainly.com/question/10874687
