One liter of a buffer composed of 1.2 m hno2 and 0.8 m nano2 is mixed with 400 ml of 0.5 m naoh. what is the new ph? assume the pka of hno2 is 3.4.

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Howell

Chemistry

6 July, 02:26

One liter of a buffer composed of 1.2 m hno2 and 0.8 m nano2 is mixed with 400 ml of 0.5 m naoh. what is the new ph? assume the pka of hno2 is 3.4.

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Jocelynn Pineda · Answer 1

Answer is: 3,4

Chemical reaction: HNO₂ + NaOH → NaNO₂ + H₂O.

c₀ (HNO₂) = 1,2 M = 1,2 mol/dm³.

c₀ (NaNO₂) = 0,8 M = 0,8 mol/dm³.

V₀ (HNO₂) = V₀ (NaNO₂) = 1 dm³ = 1 L.

c₀ (NaOH) = 0,5 M = 0,5 mol/dm³.

n₀ (HNO₂) = 1,2 mol/dm³ · 1 dm³ = 1,2 mol.

n₀ (NaNO₂) = 0,8 mol/dm³ · 1 dm³ = 0,8 mol.

V (NaOH) = 400 mL · 0,001 dm³/mL = 0,4 dm³.

n₀ (NaOH) = c₀ (NaOH) · V₀ (NaOH).

n₀ (NaOH) = 0,5 mol/dm³ · 0,4 dm³ = 0,2 mol.

n (HNO₂) = 1,2 mol - 0,2 mol = 1 mol.

n (NaNO₂) = 0,8 mol + 0,2 mol = 1 mol.

c (HNO₂) = 1 mol : 1,4 dm³ = 0,714 mol/dm³.

c (NaNO₂) = 1 mol : 1,4 dm³ = 0,714 mol/dm³.

pH = pKa + log (c (HNO₂) / c (NaNO₂)).

pH = 3,4 + log (0,714 mol/dm³ / 0,714 mol/dm³) = 3,4.