Write an equation for each acid or base showing its ionization in water, and write the equilibrium constant expression for the weak acid or base. Nitrous Acid, HNO2Ka = 4.0 x 10-4Acetic Acid, HC2H3O2Ka

HNO₂(aq) + H₂O(l)  ⇄  H₃O⁺(aq) +  NO₂⁻(aq)

Ka = [ H₃O⁺] . [NO₂⁻] / [HNO₂]

CH₃COOH(aq) + H₂O(l)  ⇄  CHCOO⁻ (aq) + H₃O⁺(aq)     Ka

Ka = [CHCOO⁻] . [H₃O⁺] / [CH₃COOH]

Remember to add the autoinization of water:

2H₂O(l) ⇄  OH⁻ (aq) + H₃O⁺(aq)              Kw

Explanation:

Nitrous acid → HNO₂ (weak acid)

The ionization in water is:

HNO₂(aq) → H⁺(aq) + NO₂⁻(aq)

Let's write the equilibrium:

HNO₂(aq) + H₂O(l)  ⇄  H₃O⁺(aq) +  NO₂⁻(aq)        Ka

Ka = [ H₃O⁺] . [NO₂⁻] / [HNO₂]

Acetic acid → CH₃COOH (weak acid)

Ionization in water: CH₃COOH(aq) → CHCOO⁻ (aq) + H⁺(aq)

Equilibrium: CH₃COOH(aq) + H₂O(l)  ⇄  CHCOO⁻ (aq) + H₃O⁺(aq)     Ka

Ka = [CHCOO⁻] . [H₃O⁺] / [CH₃COOH]

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol