Chemistry Practice Problems Ka and Kb Practice Problems Solution: Given the following Kb values, rank the conjugate ...

Solution: Given the following Kb values, rank the conjugate acids of these bases in order of increasing acid strength.HS–                Kb = 1.0 × 10–7HCOO–          Kb = 5.6 × 10–11CH3NH2         Kb = 4.4 × 10–4SO3–2             Kb = 1.6 × 10–7BrO–               Kb = 4.0 × 10–6 A. CH3NH3+ < H2S < HCOOH < HSO3– < HBrOB. CH3NH3+ < HBrO < HSO3– < H2S < HCOOHC. H2S < HSO3– < HCOOH < CH3NH3+ < HBrOD. HCOOH < H2S < HSO3– < HBrO < CH3NH3+E. H2S < HCOOH < HSO3– < CH3NH3+ < HBrO

Problem

Given the following Kb values, rank the conjugate acids of these bases in order of increasing acid strength.
HS                Kb = 1.0 × 10–7

HCOO          Kb = 5.6 × 10–11

CH3NH2         Kb = 4.4 × 10–4

SO3–2             Kb = 1.6 × 10–7

BrO               Kb = 4.0 × 10–6 

A. CH3NH3+ < H2S < HCOOH < HSO3 < HBrO

B. CH3NH3+ < HBrO < HSO3 < H2S < HCOOH

C. H2S < HSO3 < HCOOH < CH3NH3< HBrO

D. HCOOH < H2S < HSO3 < HBrO < CH3NH3+

E. H2S < HCOOH < HSO3 < CH3NH3+ < HBrO


Solution

We’re being asked to rank the conjugate acids of these bases in order of increasing acid strength given their following Kb values,


Ka represents the acid dissociation constant and it measures the strength of weak acids.

• The higher the Ka the stronger the acid and the greater the concentration of H+
• The lower the Ka, the weaker the acid and the lesser the concentration of H+


The Ka of a conjugate acid can be calculated from the Kb of a base.  Ka and Kb are connected by the autoionization constant of water (Kw) in the following equation:

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