# Chapter 16, Problem 16.133QP

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### CHEMISTRY: ATOMS FIRST VOL 1 W/CON...

14th Edition
Burdge
ISBN: 9781259327933

#### Solutions

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FindFindarrow_forward

### CHEMISTRY: ATOMS FIRST VOL 1 W/CON...

14th Edition
Burdge
ISBN: 9781259327933
Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction has to be calculated.

Concept Information:

Acid ionization constant Ka :

Acids ionize in water. Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

The ionization of a weak acid HA can be given as follows,

HA(aq)      H+(aq)+A-(aq)

The equilibrium expression for the above reaction is given below.

Ka=[H+][A-][HA]

Where,

Ka is acid ionization constant,

[H+]   is concentration of hydrogen ion

[A-]   is concentration of acid anion

[HA] is concentration of the acid

Autoionization of water:

The equation of equilibrium for autoionization of water is,

H2OH++OH-

Kw=[H+][OH-]

The equilibrium expression for water at 25oC is,

[H+][OH-]= 1×10-14

Taking negative logarithm on both sides, we get

log([H+][OH-])= -log(1×10-14)(log[H+])+(-log[OH-])= 14)

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

pH+pOH=14,at25oC

As pOH and pH are opposite scale, the total of both has to be equal to 14.

Therefore,

Kw=[H+][OH-]=1×10-14

To Calculate: The equilibrium constant for the given reaction

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