Question

Calculate the ph at of a solution of sodium hypochlorite . note that hypochlorous acid is a weak acid with a of . round your answer to decimal place.

Answer 1

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

$pK_a$ of HClO = 7.50

We know that:

$pK_a=-\log K_a$

$K_a$  of HClO = $10^{-7.50}=3.16\times 10^{-8}$

To calculate the base dissociation constant for the given acid dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant  = $3.16\times 10^{-8}$

$K_b$ = Base dissociation constant

Putting values in above equation, we get:

$10^{-14}=3.16\times 10^{-8}\times K_b\\\\K_b=\frac{10^{-14}}{3.16\times 10^{-8}}=3.16\times 10^{-7}$

The chemical equation for the reaction of hypochlorite ion with water follows:

                    $ClO^-+H_2O\rightarrow HClO+OH^-$

Initial:           0.39

At eqllm:      0.39-x                   x           x

The expression of $K_b$ for above equation follows:

$K_b=\frac{[HClO][OH^-]}{[ClO^-]}$

Putting values in above equation, we get:

$3.16\times 10^{-7}=\frac{x\times x}{(0.39-x)}\\\\x=-0.00035,0.00035$

Neglecting the negative value of 'x' because concentration cannot be negative

To calculate the pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=0.00035M$

Putting values in above equation, we get:

$pOH=-\log (0.00035)=3.6$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-3.6=10.4$

Hence, the pH of the solution is 10.4