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General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343

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BuyFindarrow_forward

General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343
Textbook Problem

A solution is prepared from 0.150 mol of formic acid and enough water to make 0.425 L of solution.

  1. a Determine the concentrations of H3O+ and HCOO in this solution.
  2. b Determine the H3O+ concentration that would be necessary to decrease the HCOO concentration above by a factor of 10. How many milliliters of 2.00 M HCl would be required to produce this solution? Consider that the solution was made by combining the HCl, the HCOOH, and enough water to make 0.425 L of solution.
  3. c Qualitatively, how can you account for the differences in the percentage dissociation of formic acid in parts a and b of this problem?

(a)

Interpretation Introduction

Interpretation:

A 0.425 L of solution is made from 0.150 mol of formic acid with addition of enough water.

The concentration of H3O+ and HCOO has to be determined

Concept Introduction:

Acid ionization constant Ka :

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

Explanation

To determine: The concentration of H3O+ and HCOO

Given data:

A 0.425 L of solution is made from 0.150 mol of formic acid with addition of enough water.

Concentration of H3O+ and HCOO :

The initial concentration of formic acid is calculated as follows,

[HCOOH]initial=0.150 mol0.425 L

Construct an equilibrium table with x as unknown concentration.

  HCOOH  +   H2O        H3O+   + HCOO
Initial (M)

0.0282

x

0.0282-x

0.00 0.00
Change (M) +x +x
Equilibrium (M) x x

Substitute into the equilibrium-constant expression

(b)

Interpretation Introduction

Interpretation:

A 0.425 L of solution is made from 0.150 mol of formic acid with addition of enough water.

Consider a 0.425 L of solution was made by addition of HCl , HCOOH and enough water. The H3O+ concentration that is needed to decrease the HCOO concentration above by a factor of 10 has to be calculated and also the amount (in milliliters) of 2.00 M HCl required has to be calculated

Concept Introduction:

Acid ionization constant Ka :

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

(c)

Interpretation Introduction

Interpretation:

A 0.425 L of solution is made from 0.150 mol of formic acid with addition of enough water.

The differences in the percentage dissociation of formic acid in parts (a) and (b) of this problem has to be accounted

Concept Introduction:

Acid ionization constant Ka :

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

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Chapter 16 Solutions

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Sect-16.4 P-16.9ESect-16.4 P-16.3CCSect-16.5 P-16.4CCSect-16.5 P-16.10ESect-16.5 P-16.11ESect-16.6 P-16.12ESect-16.6 P-16.13ESect-16.6 P-16.5CCSect-16.6 P-16.6CCSect-16.7 P-16.14ESect-16.7 P-16.15ESect-16.7 P-16.16ECh-16 P-16.1QPCh-16 P-16.2QPCh-16 P-16.3QPCh-16 P-16.4QPCh-16 P-16.5QPCh-16 P-16.6QPCh-16 P-16.7QPCh-16 P-16.8QPCh-16 P-16.9QPCh-16 P-16.10QPCh-16 P-16.11QPCh-16 P-16.12QPCh-16 P-16.13QPCh-16 P-16.14QPCh-16 P-16.15QPCh-16 P-16.16QPCh-16 P-16.17QPCh-16 P-16.18QPCh-16 P-16.19QPCh-16 P-16.20QPCh-16 P-16.21QPCh-16 P-16.22QPCh-16 P-16.23QPCh-16 P-16.24QPCh-16 P-16.25QPCh-16 P-16.26QPCh-16 P-16.27QPCh-16 P-16.28QPCh-16 P-16.29QPCh-16 P-16.30QPCh-16 P-16.31QPCh-16 P-16.32QPCh-16 P-16.33QPCh-16 P-16.34QPCh-16 P-16.35QPCh-16 P-16.36QPCh-16 P-16.37QPCh-16 P-16.38QPCh-16 P-16.39QPCh-16 P-16.40QPCh-16 P-16.41QPCh-16 P-16.42QPCh-16 P-16.43QPCh-16 P-16.44QPCh-16 P-16.45QPCh-16 P-16.46QPCh-16 P-16.47QPCh-16 P-16.48QPCh-16 P-16.49QPCh-16 P-16.50QPCh-16 P-16.51QPCh-16 P-16.52QPCh-16 P-16.53QPCh-16 P-16.54QPCh-16 P-16.55QPCh-16 P-16.56QPCh-16 P-16.57QPCh-16 P-16.58QPCh-16 P-16.59QPCh-16 P-16.60QPCh-16 P-16.61QPCh-16 P-16.62QPCh-16 P-16.63QPCh-16 P-16.64QPCh-16 P-16.65QPCh-16 P-16.66QPCh-16 P-16.67QPCh-16 P-16.68QPCh-16 P-16.69QPCh-16 P-16.70QPCh-16 P-16.71QPCh-16 P-16.72QPCh-16 P-16.73QPCh-16 P-16.74QPCh-16 P-16.75QPCh-16 P-16.76QPCh-16 P-16.77QPCh-16 P-16.78QPCh-16 P-16.79QPCh-16 P-16.80QPCh-16 P-16.81QPCh-16 P-16.82QPCh-16 P-16.83QPCh-16 P-16.84QPCh-16 P-16.85QPCh-16 P-16.86QPCh-16 P-16.87QPCh-16 P-16.88QPCh-16 P-16.89QPCh-16 P-16.90QPCh-16 P-16.91QPCh-16 P-16.92QPCh-16 P-16.93QPCh-16 P-16.94QPCh-16 P-16.95QPCh-16 P-16.96QPCh-16 P-16.97QPCh-16 P-16.98QPCh-16 P-16.99QPCh-16 P-16.100QPCh-16 P-16.101QPCh-16 P-16.102QPCh-16 P-16.103QPCh-16 P-16.104QPCh-16 P-16.105QPCh-16 P-16.106QPCh-16 P-16.107QPCh-16 P-16.108QPCh-16 P-16.109QPCh-16 P-16.110QPCh-16 P-16.111QPCh-16 P-16.112QPCh-16 P-16.113QPCh-16 P-16.114QPCh-16 P-16.115QPCh-16 P-16.116QPCh-16 P-16.117QPCh-16 P-16.118QPCh-16 P-16.119QPCh-16 P-16.120QPCh-16 P-16.121QPCh-16 P-16.122QPCh-16 P-16.123QPCh-16 P-16.124QPCh-16 P-16.125QPCh-16 P-16.126QPCh-16 P-16.127QPCh-16 P-16.128QPCh-16 P-16.129QPCh-16 P-16.130QPCh-16 P-16.131QPCh-16 P-16.132QPCh-16 P-16.133QPCh-16 P-16.134QPCh-16 P-16.135QPCh-16 P-16.136QPCh-16 P-16.137QPCh-16 P-16.138QPCh-16 P-16.139QPCh-16 P-16.140QPCh-16 P-16.141QPCh-16 P-16.142QPCh-16 P-16.143QPCh-16 P-16.144QPCh-16 P-16.145QPCh-16 P-16.146QPCh-16 P-16.147QPCh-16 P-16.148QPCh-16 P-16.149QPCh-16 P-16.150QPCh-16 P-16.151QPCh-16 P-16.152QPCh-16 P-16.153QPCh-16 P-16.154QPCh-16 P-16.155QPCh-16 P-16.156QPCh-16 P-16.157QP

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