(1) h a ( a q) + h 2 o ( l) ⇌ h 3 o ( a q) + + a ( a q) −. 4.5 ∗ 10 − 4 = x 2 / ( 0.2 − x) rearrange: X ≈ 0.009 p h = − l o g ( 10) p h ≈ 2.05. Write the balanced dissociation equation for the weak acid. Part a answer o2 (aq)⇌h+ (aq)+no−2 (aq) part b by using the value of ka, calculate δ g ∘ for the dissociation of nitrous acid in aqueous solution. Share improve this answer follow Ka is the equilibrium constant for the dissociation. Web the value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4 part a write the chemical equation for the equilibrium that corresponds to ka. Part b answer δg∘ = 19.1 kj Web a large k a value indicates a stronger acid (more of the acid dissociates) and small k a value indicates a weaker acid (less of the acid dissociates).
Web the ka value, also known as the acid dissociation equilibrium constant, is a measure of the acidity of a solution. Web a large k a value indicates a stronger acid (more of the acid dissociates) and small k a value indicates a weaker acid (less of the acid dissociates). 3) calculate the ph of 0.50 m nano2 (sodium nitrite). Create an initial change equilibrium (ice) table for the. How do you write ka expression? Part b answer δg∘ = 19.1 kj Web steps in determining the ka of a weak acid from ph step 1: For a chemical equation of the form (3) h a + h 2 o ⇋ h 3 o + + a − k a is express as (4) k a = [ h 3 o +] [ a −] [ h a] where h a is the undissociated acid and a − is the conjugate base of the acid. The ka value of an acid is related to its ph. R = gas constant = 8.314 j/k. Web solved what is the equilibrium constant expression (ka) for | chegg.com.
