Question

This FRQ is kinda confusing me, could someone help me please :)

127.Hypochlorous acid, HOCI, is a weak acid in water. The K, expression for HOCI is shown above Write a chemical equation showing how HOCI behaves as an acid in water a. Calculate the pH ofa 0.175 M solution of HOCI b. Write the net ionic equation for the reaction between the weak acid HOCI(aq) and the strong base NaOH(aq) In an experiment, 20.00 mL of 0.175 M HOCI(aq) is placed in a flask and titrated with 6.55 mL of 0.435 M NaOH(aq) d. Calculate the number of moles of NaOH(aq) added i) Calculate [H,O' in the flask after the NaOH(aq) has been added. Calculate [OH] in the flask after the NaOH(aq) has been added. 128.NH (aq)H0(1)-NH (aq)+OH (aq) In aqueous solution, ammonia reacts as represented above. In 0.0180 M NH(aq) at 25°C, the hydroxide ion concentration, [OH], is 5.60 x 10 M. In answering the following, assume that temperature is constant at 25°C and that volumes are additive Write the equilibrium-constant expression for the reaction represented above a. Determine the pH of 0.01 80 MNH (aq) b. Determine the value of the base ionization constant, K, for NH,(aq) c. Determine the percent ionization of NH, in 0.0180 M NH(aq) d. In an experiment, a 20.0 mL sample of 0.0 1 80 M NH (aq) was placed in a flask and titrated to the equivalence point and beyond using 0.0120 M HCI(aq) Determine the volume of 0.0120 M HCl(aq) that was added to reach the equivalence point. i. Determine the pH of the solution in the flask after a total of 15.0 mL of 0.0120 M HCl(aq) was added ii. Determine the pH of the solution in the flask after a total of 40.0 mL of 0.0120 M HCl(aq) was added

Answer 1

Ans 127

Part a

The balanced reaction of HOCl in water

HOCl (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + ClO^- (aq)    

Ka = 2.9 x 10^-8

Part b

Concentration of HOCl = 0.175 M

The balanced reaction with ICE TABLE

HOCl (aq) + H2O (l) ⇌ H3O+ (aq) + ClO- (aq)

I 0.175 0 0

C -x +x +x

E 0.175-x x x

Equilibrium constant expression of the reaction

Ka = [H3O+] [ClO-] / [HOCl]

3.2 * 10^-8 = (x) (x) / (0.175 - x)

x <<< 0.175

Then, 3.2 * 10^-8 = (x^2) / (0.175 )

x = 7.48*10^-5 = [H3O+]

pH = - log[H3O+]

= - log (7.48*10^-5) = 4.13

Part c

The balanced reaction

NaOH(aq) + HOCl (l) = NaOCl(aq) + H2O (l)

Total ionic equation

Na+(aq) + OH-(aq) + HOCl (l) = Na+(aq) + OCl-(aq) + H2O (l)

Net ionic equation

OH-(aq) + HOCl (l) = OCl-(aq) + H2O (l)

Part d

(i)

Moles of NaOH = molarity of NaOH x volume of NaOH

= 0.435 Mol/L x 6.55 mL x 1L/1000 mL

= 0.00284925 mol

= 2.85*10^-3 mol

ii) moles of HOCl = molarity x volume

= 0.175 M x 20 mL x 1L/1000 mL

= 0.00350 mol

= 3.50*10^-3 mol

Moles of HOCl in excess = 3.50*10^-3 - 2.85*10^-3

= 6.50*10^-4 mol

Total volume V = 20 + 6.55 = 26.55 mL x 1L/1000 mL

= 2.655 *10^-2 L

Concentration of HOCl = 6.50*10^-4 mol / 2.655 *10^-2 L

= 2.45 * 10^-2 M

Moles of OCl- = moles of NaOH reacted = 2.85*10^-3 mol

Concentration of OCl- = 2.85*10^-3 mol/2.655 *10^-2 L

= 0.107 M

pH = pKa + log [ OCl-] / [HOCl]

= 7.49 + log (0.107/2.45 * 10^-2)

= 8.14

[H+] = 10^-pH = 10^-8.14

= 7.24 * 10^-9 M

iii)

[OH-] = (10^-14 / 7.24 * 10^-9)

= 1.38*10^-6 M