Answer:
0.0905 M
Explanation:
Let's consider the neutralization reaction between H2SO4 and KOH.
H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O
22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:
0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol
The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol
1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:
M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M
Answer:
Percentage dissociated = 0.41%
Explanation:
The chemical equation for the reaction is:

The ICE table is then shown as:

Initial (M) 1.8 0 0
Change (M) - x + x + x
Equilibrium (M) (1.8 -x) x x
![K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}](https://tex.z-dn.net/?f=K_a%20%20%3D%20%5Cfrac%7B%5BC_3H_6ClCO%5E-_2%5D%5BH%5E%2B%5D%7D%7B%5BC_3H_6ClCO_2H%5D%7D)
where ;


Since the value for
is infinitesimally small; then 1.8 - x ≅ 1.8
Then;




Dissociated form of 4-chlorobutanoic acid = 
Percentage dissociated = 
Percentage dissociated = 
Percentage dissociated = 0.4096
Percentage dissociated = 0.41% (to two significant digits)
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