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Katen [24]
3 years ago
11

When 0.040 mol of propionic acid, c2h5co2h, is dissolved in 750 ml of water, the equilibrium concentration of h3o+ ions is measu

red to be 1.84 x 10-3 m. what is ka for this acid?
Chemistry
1 answer:
WITCHER [35]3 years ago
3 0
Answer is: Ka for propanoic acid is 6,57·10⁻⁵.
Chemical reaction: C₂H₅COOH(aq) + H₂O(l) ⇄ C₂H₅COO⁻(aq) + H₃O⁺(aq).
n(C₂H₅COOH) = 0,04 mol.
V(C₂H₅COOH) = 750 mL = 0,75 L.
c(C₂H₅COOH) = 0,04 mol ÷ 0,75 L.
c(C₂H₅COOH) = 0,053 mol/L = 0,053 M.
[C₂H₅COO⁻] = [H₃O⁺] = 1,84·10⁻³ M = 0,00184 M.<span>
[HCN] = 0,053 M - 0,00184 M = 0,0515 M.
Ka = [</span>C₂H₅COO⁻] · [H₃O⁺] / [C₂H₅COOH].
Ka = (0,00184 M)² / 0,0515 M.
Ka = 6,57·10⁻⁵.
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the equilibrium concentration of H₂(g) at 700°C =  0.00193 mol/L

0.00193 mol/L

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numbers of moles of H₂S = 0.59 moles

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Equilibrium constant  = 9.30 × 10⁻⁸

The equation for the reaction is given as :

2H₂S    ⇄   2H₂(g)  +  S₂(g)

The initial concentration of H₂S =

The initial concentration of H₂S =

= 0.1966 mol/L

The ICE table is shown be as :

                           2H₂S                ⇄         2H₂(g)        +        S₂(g)

Initial                    0.9166                           0                           0

Change                 -2 x                               +2 x                      + x

Equilibrium          (0.9166 - 2x)                   2x                         x

     

(since 2x < 0.1966 if solved through quadratic equation)

The equilibrium concentration for H₂(g) = 2x

∴

= 0.00193 mol/L

Thus, the equilibrium concentration of H₂(g) at 700°C =  0.00193 mol/L

To know more about equilibrium concentration

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