BuyFindarrow_forward

General Chemistry - Standalone boo...

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

Solutions

Chapter
Section
BuyFindarrow_forward

General Chemistry - Standalone boo...

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

A 50.0-mL sample of a 0.100 M solution of NaCN is titrated by 0.200 M HCl. Kb for CN is 2.0 × 10−5. Calculate the pH of the solution: a prior to the start of the titration; b after the addition of 15.0 mL of 0.200 M HCl; c at the equivalence point; d after the addition of 30.0 mL of 0.200 M HCl.

(a)

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of NaCN   with hydrochloric acid has to be calculated.

  1. (a) Prior to the start of the titration
  2. (b) After the addition of 15 mL of 0.200 M HCl
  3. (c) At the equivalence point
  4. (d) After the addition of 30.0 mL of 0.200 M HCl

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH] concentration.

pOH=-log[OH]

Relationship between pH  and pOH:

pH + pOH = 14

Explanation

To Calculate: The pH prior to the start of the titration

Given data:

The volume of NaCN = 50.0 mL

The concentration of NaCN = 0.100 M

The concentration of hydrochloric acid = 0.200 M

The Kb value for CN is 2.0×105

pH prior to the start of the titration

Construct an equilibrium table for the hydrolysis of cyanide ion from NaCN as follows,

     CN-  +   H2O        HCN    +   OH
Initial (M)

0.100

x

0.100-x

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

The Kb value for CN is 2.0×105

Now substitute equilibrium concentrations into the equilibrium-constant expression.

      Kb =(x)2(0

(b)

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of NaCN   with hydrochloric acid has to be calculated.

  1. (a) Prior to the start of the titration
  2. (b) After the addition of 15 mL of 0.200 M HCl
  3. (c) At the equivalence point
  4. (d) After the addition of 30.0 mL of 0.200 M HCl

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH] concentration.

pOH=-log[OH]

Relationship between pH  and pOH:

pH + pOH = 14

(c)

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of NaCN   with hydrochloric acid has to be calculated.

  1. (a) Prior to the start of the titration
  2. (b) After the addition of 15 mL of 0.200 M HCl
  3. (c) At the equivalence point
  4. (d) After the addition of 30.0 mL of 0.200 M HCl

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH] concentration.

pOH=-log[OH]

Relationship between pH  and pOH:

pH + pOH = 14

(d)

Interpretation Introduction

Interpretation:

The pH of the given points of the titration of NaCN   with hydrochloric acid has to be calculated.

  1. (a) Prior to the start of the titration
  2. (b) After the addition of 15 mL of 0.200 M HCl
  3. (c) At the equivalence point
  4. (d) After the addition of 30.0 mL of 0.200 M HCl

Concept Introduction:

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH] concentration.

pOH=-log[OH]

Relationship between pH  and pOH:

pH + pOH = 14

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Chapter 16 Solutions

Show all chapter solutions add
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

Additional Science Solutions

Find more solutions based on key concepts

Show solutions add

LDL, a class of lipoprotein, delivers triglycerides and cholesterol from the liver to the bodys tissues. T F

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

Why are interstellar lines so narrow?

Horizons: Exploring the Universe (MindTap Course List)

A thin, square, conducting plate 50.0 cm on a side lies in the xy plane. A total charge of 4.00 108 C is place...

Physics for Scientists and Engineers, Technology Update (No access codes included)