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)?

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

for sodium formate is

Given that:

of formic acid =

And,

So,

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:

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

answer: mercury

explanation: