Asolution of an unknown acid had a ph of 3.70. titration of a 25.0 ml aliquot of the acid solution required 21.7 ml of 0.104 m sodium hydroxide for complete reaction. assuming that the acid is monoprotic, what is its ionization constant

Ionization constant for acid = Ka

let's assume that monoprotic acid is HA

the reaction between HA and NaOH,
      HA(aq) + NaOH(aq) → NaA(aq) + H₂O(l)

the stoichiometric ratio between HA and NaOH is 1 : 1
Hence,
    Moles of NaOH added = reacted moles of HA in 25.00 mL

Moles of NaOH added = concentarion x volume added
                                    = 0.104 mol L⁻¹ x 21.70 x 10⁻³ L
                                    
reacted moles of HA in 25.00 mL = 0.104 mol L⁻¹ x 21.70 x 10⁻³ L
hence, the initial [HA]         = moles / volume
                                           = (0.104 mol L⁻¹ x 21.70 x 10⁻³ L)  / 25.00 x 10⁻³ L
                                           = 0.090 M

According to the pH, the molar solubility = [H⁺(aq)] = X

pH    = -log[H⁺(aq)] 
3.70 = -log[H⁺(aq)] 
[H⁺(aq)]  = 1.995 x 10⁻⁴ M
X   = 1.995 x 10⁻⁴ M
       
 at equilibrium,
                      HA(aq) ⇄ H⁺(aq) + A⁻(aq)
Initial              0.090
Change           -X               +X         +X
Equilibrium    0.090 - X        X         X

Ka = [H⁺(aq)] [ A⁻(aq)] / [HA(aq)]
     = X x X / (0.090 - X)
     = (1.995 x 10⁻⁴ M)² / (0.090 - 1.995 x 10⁻⁴) M
     = 4.432 x 10⁻⁷ M

Sorry i dont understand the question can you rephase it?