Question

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 1

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa    ⇒     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:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

pH = 14 - 4.64 = 9.36