# A 50.0-mL sample of a 0.100 M solution of NaCN is titrated by 0.200 M HCl. K b 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.

### General Chemistry - Standalone boo...

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

### General Chemistry - Standalone boo...

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

#### Solutions

Chapter
Section
Chapter 16, Problem 16.91QP
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.

Expert Solution

(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 of Solution

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

Expert Solution

(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

Expert Solution

(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

Expert Solution

(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

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