Answer: The pH of the solution is 6.81
Explanation:
The chemical equation for the ionization of water follows:
[tex]H_2O\rightleftharpoons H^++OH^-[/tex]
The expression of [tex]K_w[/tex] for above equation, we get:
[tex]K_w=[H^+]\times [OH^-][/tex]
We are given:
[tex]K_w=2.4\times 10^{-14}[/tex]
[tex][H^+]=[OH^-]=x[/tex]
Putting values in above equation, we get:
[tex]2.4\times 10^{-14}=x\times x\\\\x=1.55\times 10^{-7}M[/tex]
[tex]pH=-\log[H^+][/tex]
We are given:
[tex][H^+]=1.55\times 10^{-7}M[/tex]
Putting values in above equation, we get:
[tex]pH=-\log(1.55\times 10^{-7})\\\\pH=6.81[/tex]
Hence, the pH of the solution is 6.81
The reaction is:
For neutral solution,
then,
→ [tex]K_w = [H^+] [OH^-] = [H^+]^2[/tex]
→ [tex][H^+]^2 = 2.4\times 10^{-14}[/tex]
[tex][H^+] = 1.5\times 10^{-7}[/tex]
We know,
→ [tex]pH = -log[H^+][/tex]
By substituting the value of "[tex][H^+][/tex]", we get
[tex]= -log(1.5\times 10^{-7})[/tex]
[tex]= 6.82[/tex]
Thus the above answer is right.
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