The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

Answer:9.36Explanation:Sodium formate is the conjugate base of formic acid. Also,[tex]K_a\times K_b=K_w[/tex][tex]K_b[/tex] for sodium formate is [tex]K_b=\frac {K_w}{K_a}[/tex]Given that:[tex]K_a[/tex] of formic acid = [tex]1.8\times 10^{-4}[/tex]And, [tex]K_w=10^{-14}[/tex]So, [tex]K_b=\frac {10^{-14}}{1.8\times 10^{-4}}[/tex][tex]K_b=5.5556\times 10^{-11}[/tex]Concentration = 0.35 MHCOONa    ⇒     Na⁺ +    HCOO⁻Consider the ICE take for the formate  ion as:                                    HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻At t=0                              0.35                            -              -At t =equilibrium           (0.35-x)                          x           x            The expression for dissociation constant of sodium formate is:[tex]K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}[/tex][tex]5.5556\times 10^{-11}=\frac {x^2}{0.35-x}[/tex]Solving for x, we get:x = 0.44×10⁻⁵  MpOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64pH + pOH = 14So, pH = 14 - 4.64 = 9.36