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
If we dissolve salt in water they will reduce the intermolecular forces between water molecule and this will decrease the surface tension.
Surface tension is due to cohesive forces (the forces between molecules of same substance) hence as cohesive forces decreases the surface tension also decreases
