Answer:
Ka = 1.52 E-5
Explanation:
- CH3-(CH2)2-COOH ↔ CH3(CH2)2COO- + H3O+
⇒ Ka = [H3O+][CH3)CH2)2COO-] / [CH3(CH2)2COOH]
mass balance:
⇒<em> C</em> CH3(CH2)2COOH = [CH3(CH2)2COO-] + [CH3(CH2)2COOH] = 1.0 M
charge balance:
⇒ [H3O+] = [CH3(CH2)2COO-]
⇒ Ka = [H3O+]²/(1 - [H3O+])
∴ pH = 2.41 = - Log [H3O+]
⇒ [H3O+] = 3.89 E-3 M
⇒ Ka = (3.89 E-3)² / ( 1 - 3.89 E-3 )
⇒ Ka = 1.519 E-5
For balancing acidic solutions, we would need to add H+ ions to the correct side of the equation to balance the total number of atoms and the overall charge.
The third is one: controlled burn
<span>(1) CH3CHCHCH3.................</span>
