Failed to connect to MySQL: Access denied for user 'examnext_online'@'localhost' (using password: YES) 20 mL of a weak acid HX is titrated against 0.1 M NaOH. At the point of half equivalence the pH of solution is 5.7. hence, `K_a` of acid is - Sarthaks eConnect

20 mL of a weak acid HX is titrated against 0.1 M NaOH. At the point of half equivalence the pH of solution is 5.7. hence, `K_a` of acid is

20 mL of a weak acid HX is titrated against 0.1 M NaOH. At the point of half equivalence the pH of solution is 5.7. hence, `K_a` of acid is

A.

`7.0×10^-5`

B.

`2.0×10^-5`

C.

`7.0×10^-4`

D.

`2.0×10^-6`

Solution

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Best answer

At the point of half neutralization of mixture will be an acid buffer with [salt]/[acid]=1

`pH = pK_a + log1`

pKa = 5.7

`=>K_a = 2.0×10^(–6)`