# Chapter 16, Problem 16.97QP

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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 pH of a 0.91MC2H5NH3I solution has to be calculated

Concept Information:

Acid ionization constant Ka :

The equilibrium expression for the reaction HA(aq)+H2O   H3O+(aq)+A-(aq) is given below.

Ka=[H3O+][A-][HA]

Where Ka is acid ionization constant, [H3O+]   is concentration of hydronium ion, [A-]   is concentration of acid anion, [HA] is concentration of the acid

Base ionization constant Kb

The equilibrium expression for the ionization of weak base B will be,

B(aq)+H2O(l)HB+(aq)+OH-(aq)

Kb=[HB+][OH-][B]

Where Kb is base ionization constant, [OH] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base

Relationship between Ka and Kb

Ka×Kb=Kw

Where,

Kw is autoionization of water.

The autoionization of water:Kw=[H+][OH-]=1.0×1014

Relationship between pH and pOH:

For calculation of pH, the following formula is used,

pH=log[H+]

For calculation of pOH, the following formula is used,

pOH=log[OH-]

From definition of pH and pOH, we get at room temperature,

pH+pOH=14

To Calculate: The pH of a 0.91MC2H5NH3I solution

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