Question

A solution was prepared by dissolving 0.200 moles of sodium hypoch lorite (NaClO) in water to a final volume of 1.00 Liter. Ka 2.96x10 for hypochlorous acid (HCIO). Show equations and work steps for these calculations. 1) Calculate Kb of sodium hypochlorite. 2) Calculate the pH of the NaClO solution. 3) Calculate the pH if 10.0 mL of 1.00 M HCl is added to 100.0 mL of the original NACIO solution. 4) Calculate the pH if 20.0 mL of 1.00 M HCl is added to 10o.0 mL of the original NaCIO solution.

Answer 1

1) HClO and ClO ^- is conjugate acid base pair.

For conjugate acid-base pair , we have K a $\times$ K b = K w

$\therefore$ K b of ClO ^- = K w / K a of HClO = ( 1.0 $\times$ 10 ^-14 ) / ( 2.96 $\times$ 10 ^-08 ) = 3.38 $\times$   $\times$ 10 ^-07

2) [NaClO] = No. of moles of NaClO / volume of solution in L

[NaClO ] = 0.200 mol / 1.00 L = 0.200 M

Consider dissociation of ClO ^- in water.

ClO ^- (aq) + H₂O (l) $\rightleftharpoons$ HClO (aq) + OH ^- (aq)

For above reaction, K b = [HClO] [OH ^-] / [ClO ^- ] = 3.38 $\times$   $\times$ 10 ^-07

Let's use ICE table.

Concentration (M)	ClO -	HClO	OH -
Initial	0.200
Change	- A	+A	+A
Equilibrium	0.200 -A	A	A

From above equilibrium concentrations , we get

K b = ( A) (A) / 0.200 - A =  3.38 $\times$   $\times$ 10 ^-07

A ² / 0.200 - A =  3.38 $\times$   $\times$ 10 ^-07

Assume A is very small as compared to 0.200. Therefore, we can write 0.200 -A $\approx$ 0.200.

A ² / 0.200 =  3.38  $\times$ 10 ^-07

A ² = 0.200 $\times$   3.38  $\times$ 10 ^-07

A ² = 6.76 $\times$ 10 ^-08

Taking square root on both sides , we get A = 2.6 $\times$ 10 ^-04 M = [OH ^-]

We have relation, [H ⁺ ] [OH ^-] = 10 ^-14

Therefore, [H ⁺ ] = 10 ^-14 / [H ⁺ ] = 10 ^-14 / 2.6 $\times$ 10 ^-04 = 3.846 $\times$ 10 ^-11 M

$\therefore$ pH = -log [H ⁺ ] = -log 3.846 $\times$ 10 ^-11 = 10.4

3) Consider reaction , HCl + NaClO $\rightarrow$ NaCl + HClO

No. of moles of HCl = [HCl] $\times$ volume of solution in L = 1.00 mol / L $\times$ 0.01 L = 0.0100 mol

No. of moles of NaClO = [NaClO] $\times$ volume of solution in L = 0.200 mol / L $\times$ 0.100 L = 0.02 mol

Moles of excess NaClO = 0.02 - 0.01 = 0.01 mol

No. of moles of HClO produced = No. of moles of HCl added = 0.0100 mol

Volume of solution = volume of NaClO + volume of HCl = 200 + 10 = 210 ml = 0.210 L

[NaClO] = 0.01 mol / 0.210 L = 0.0476 M

[HClO] = 0.01 mol / 0.210 L = 0.0476 M

Solution contain weak acid HClO and its conjugate base ClO ^-. Hence it acts as a buffer solution .

pH of buffer is calculated by using Henderson's equation.

pH = pKa + log [NaClO] / [HClO]

pH = - log K a + log [NaClO] / [HClO]

pH = -log ( 2.96 $\times$ 10 ^-08 ) + log 0.0476 / 0.0476

pH = 7.529 + log 1

pH = 7.529 + 0

pH = 7.529

4) No. of moles of HCl = [HCl] $\times$ volume of solution in L = 1.00 mol / L $\times$ 0.02 L = 0.0200 mol

No. of moles of NaClO = [NaClO] $\times$ volume of solution in L = 0.200 mol / L $\times$ 0.100 L = 0.0200 mol

We get , No. of moles of acid = No. of moles of Base = 0.0200 mol . Hence, all acid is neutralized by base. So pH of solution will be due to dissociation of HClO in water.

No. of moles of HClO produced = No. of moles of HCl added = 0.0200 mol

Volume of solution = volume of NaClO + volume of HCl = 200 + 20 = 220 ml = 0.220 L

[HClO] = 0.0200 mol / 0.220 L = 0.0909 M

Consider reaction, HClO (aq) + H₂O (l) $\rightleftharpoons$ H₃O ⁺ (aq) + ClO ^- (aq)

For above reaction , K a = [ H₃O ⁺] [ClO ^-] / [HClO] = 2.96 $\times$ 10 ^-08

Let's use ICE table.

Concentration (M)	HClO	H3O +	ClO -
Initial	0.0909
Change	- A	+A	+A
Equilibrium	0.0909 -A	A	A

From above equilibrium concentrations , we get

K b = ( A) (A) / 0.0909 - A = 2.96 $\times$   $\times$ 10 ^-08

A ² / 0.0909 - A =  2.96 $\times$   $\times$ 10 ^-08

Assume A is very small as compared to 0.0909. Therefore, we can write 0.0909 -A $\approx$ 0.0909.

A ² / 0.0909 =   2.96 $\times$   $\times$ 10 ^-08

A ² = 0.0909 $\times$ 2.96 $\times$   $\times$ 10 ^-08

A ² = 2.69 $\times$ 10 ^-09

Taking square root on both sides , we get A =5.19 $\times$ 10 ^-05 M = [H₃O ⁺]

We have, pH = -log   [H₃O ⁺] = - log 5.19 $\times$ 10 ^-05 = 4.28