Respuesta :
The question is incomplete, here is the complete question:
Calculate the pH at 25°C of a 0.39 M solution of sodium hypochlorite NaClO. Note that hypochlorous acid HClO is a weak acid with a pKa of 7.50. Round your answer to 1 decimal place.
Answer: The pH of the solution is 10.4
Explanation:
We are given:
Molarity of sodium hypochlorite = 0.39 M
[tex]pK_a[/tex] of HClO = 7.50
We know that:
[tex]pK_a=-\log K_a[/tex]
[tex]K_a[/tex] of HClO = [tex]10^{-7.50}=3.16\times 10^{-8}[/tex]
To calculate the base dissociation constant for the given acid dissociation constant, we use the equation:
[tex]K_w=K_b\times K_a[/tex]
where,
[tex]K_w[/tex] = Ionic product of water = [tex]10^{-14}[/tex]
[tex]K_a[/tex] = Acid dissociation constant = [tex]3.16\times 10^{-8}[/tex]
[tex]K_b[/tex] = Base dissociation constant
Putting values in above equation, we get:
[tex]10^{-14}=3.16\times 10^{-8}\times K_b\\\\K_b=\frac{10^{-14}}{3.16\times 10^{-8}}=3.16\times 10^{-7}[/tex]
The chemical equation for the reaction of hypochlorite ion with water follows:
[tex]ClO^-+H_2O\rightarrow HClO+OH^-[/tex]
Initial: 0.39
At eqllm: 0.39-x x x
The expression of [tex]K_b[/tex] for above equation follows:
[tex]K_b=\frac{[HClO][OH^-]}{[ClO^-]}[/tex]
Putting values in above equation, we get:
[tex]3.16\times 10^{-7}=\frac{x\times x}{(0.39-x)}\\\\x=-0.00035,0.00035[/tex]
Neglecting the negative value of 'x' because concentration cannot be negative
To calculate the pOH of the solution, we use the equation:
[tex]pOH=-\log[OH^-][/tex]
We are given:
[tex][OH^-]=0.00035M[/tex]
Putting values in above equation, we get:
[tex]pOH=-\log (0.00035)=3.6[/tex]
To calculate pH of the solution, we use the equation:
[tex]pH+pOH=14\\pH=14-3.6=10.4[/tex]
Hence, the pH of the solution is 10.4
