In an exactly.200m solution of benzoic acid, a monoprotic acid, the [h+] = 3.55 × 10–3m. what is the ka for benzoic acid? 8.88 × 10–2 1.77 × 10–2 6.3 × 10–5 7.10 × 10–4

Answer : Option C) 6.3 X .

Explanation : Given is the concentration of benzoic acid is 0.200 M, now only one monoprotic acid ion is getting separated from it [.

So, the reaction will be as,

   ⇔  + .

We can use the ICE method for calculation,
 The molar concentration of is 0.200 M and the  [ is 3.55 X .

So here on reactant side we get, (0.2 - 3.55 X ) = 0.196
 on the product it remains as 3.55 X .

So,  = (3.55 X ) X (3.55 X ) / 0.196

We will get as 6.42 X  which is close to 6.3 X .