The value of pH for 0.100 M Fe ( NO 3 ) 3 is to be calculated. Concept introduction: The potential of H + ions is called as pH . The value of pH is less than 7 for acidic solutions and greater than 7 for basic solutions. For neutral solution the value of pH is 7 . To determine: The value of pH for 0.100 M Fe ( NO 3 ) 3 .

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Answer 1

Question

Chapter 17, Problem 17.51QP

Interpretation Introduction

Interpretation: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is to be calculated.

Concept introduction: The potential of $H+$ ions is called as $pH$ . The value of $pH$ is less than $7$ for acidic solutions and greater than $7$ for basic solutions. For neutral solution the value of $pH$ is $7$ .

To determine: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ .

Expert Solution & Answer

Answer to Problem 17.51QP

The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is $1.72_$ .

Explanation of Solution

The concentration of $Fe(NO3)3$ is $0.100 M$ .

The chemical equation for the dissociation of $Fe(NO3)3$ in water is,

$Fe(NO3)3+3H2O→Fe(OH)3+3HNO3$

The net ionic equation is,

$Fe3++3H2O→Fe(OH)3+3H+$

The $Ka$ of $Fe(NO3)3$ is $3.00×10−3$ .

The ICE table for the dissociation of $Fe(NO3)3$ is,

$Fe3+(aq)→Fe(OH)3(aq)+H+(aq)I0.10000C−xxxE0.100−xxx$

The concentrations of both $Fe(OH)3$ and $H+$ are equal and assumed to be $x$ and is calculated by the equilibrium constant expression,

$Ka=[Fe(OH)3][H+][Fe3+]$

Substitute the values of dissociation constant and concentration of base in the above expression to calculate the concentration of $H+$ ions.

$3.00×10−3=x20.100−xx2=3.00×10−3(0.100−x)x2=3.00×10−4−3.00×10−3x0=x2+3.00×10−3x−3.00×10−4$

Simplify the above equation by the formula,

$x=−b±b2−4ac2a$

Substitute the values of $a$ and $b$ in the above formula to calculate the value of $x$ .

$x=−(3.00×10−3)±(3.00×10−3)2−4×1×(−3.00×10−4)2×1=−3.00×10−3±9.0×10−6+1.2×10−32=−3.00×10−3+0.03472=0.01885 M$

Hence, the concentration of $H+$ is $0.01885 M$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above formula to calculate the value of $pH$ .

$pH=−log(0.01885)=1.72_$

Hence, the $pH$ of a solution is $1.72_$ .

Conclusion

The $pH$ of a $Fe(NO3)3$ solution is $1.72_$ .

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Answer 2

Question

Chapter 17, Problem 17.51QP

Interpretation Introduction

Interpretation: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is to be calculated.

Concept introduction: The potential of $H+$ ions is called as $pH$ . The value of $pH$ is less than $7$ for acidic solutions and greater than $7$ for basic solutions. For neutral solution the value of $pH$ is $7$ .

To determine: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ .

Expert Solution & Answer

Answer to Problem 17.51QP

The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is $1.72_$ .

Explanation of Solution

The concentration of $Fe(NO3)3$ is $0.100 M$ .

The chemical equation for the dissociation of $Fe(NO3)3$ in water is,

$Fe(NO3)3+3H2O→Fe(OH)3+3HNO3$

The net ionic equation is,

$Fe3++3H2O→Fe(OH)3+3H+$

The $Ka$ of $Fe(NO3)3$ is $3.00×10−3$ .

The ICE table for the dissociation of $Fe(NO3)3$ is,

$Fe3+(aq)→Fe(OH)3(aq)+H+(aq)I0.10000C−xxxE0.100−xxx$

The concentrations of both $Fe(OH)3$ and $H+$ are equal and assumed to be $x$ and is calculated by the equilibrium constant expression,

$Ka=[Fe(OH)3][H+][Fe3+]$

Substitute the values of dissociation constant and concentration of base in the above expression to calculate the concentration of $H+$ ions.

$3.00×10−3=x20.100−xx2=3.00×10−3(0.100−x)x2=3.00×10−4−3.00×10−3x0=x2+3.00×10−3x−3.00×10−4$

Simplify the above equation by the formula,

$x=−b±b2−4ac2a$

Substitute the values of $a$ and $b$ in the above formula to calculate the value of $x$ .

$x=−(3.00×10−3)±(3.00×10−3)2−4×1×(−3.00×10−4)2×1=−3.00×10−3±9.0×10−6+1.2×10−32=−3.00×10−3+0.03472=0.01885 M$

Hence, the concentration of $H+$ is $0.01885 M$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above formula to calculate the value of $pH$ .

$pH=−log(0.01885)=1.72_$

Hence, the $pH$ of a solution is $1.72_$ .

Conclusion

The $pH$ of a $Fe(NO3)3$ solution is $1.72_$ .

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Answer 3

Question

Chapter 17, Problem 17.51QP

Interpretation Introduction

Interpretation: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is to be calculated.

Concept introduction: The potential of $H+$ ions is called as $pH$ . The value of $pH$ is less than $7$ for acidic solutions and greater than $7$ for basic solutions. For neutral solution the value of $pH$ is $7$ .

To determine: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ .

Expert Solution & Answer

Answer to Problem 17.51QP

The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is $1.72_$ .

Explanation of Solution

The concentration of $Fe(NO3)3$ is $0.100 M$ .

The chemical equation for the dissociation of $Fe(NO3)3$ in water is,

$Fe(NO3)3+3H2O→Fe(OH)3+3HNO3$

The net ionic equation is,

$Fe3++3H2O→Fe(OH)3+3H+$

The $Ka$ of $Fe(NO3)3$ is $3.00×10−3$ .

The ICE table for the dissociation of $Fe(NO3)3$ is,

$Fe3+(aq)→Fe(OH)3(aq)+H+(aq)I0.10000C−xxxE0.100−xxx$

The concentrations of both $Fe(OH)3$ and $H+$ are equal and assumed to be $x$ and is calculated by the equilibrium constant expression,

$Ka=[Fe(OH)3][H+][Fe3+]$

Substitute the values of dissociation constant and concentration of base in the above expression to calculate the concentration of $H+$ ions.

$3.00×10−3=x20.100−xx2=3.00×10−3(0.100−x)x2=3.00×10−4−3.00×10−3x0=x2+3.00×10−3x−3.00×10−4$

Simplify the above equation by the formula,

$x=−b±b2−4ac2a$

Substitute the values of $a$ and $b$ in the above formula to calculate the value of $x$ .

$x=−(3.00×10−3)±(3.00×10−3)2−4×1×(−3.00×10−4)2×1=−3.00×10−3±9.0×10−6+1.2×10−32=−3.00×10−3+0.03472=0.01885 M$

Hence, the concentration of $H+$ is $0.01885 M$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above formula to calculate the value of $pH$ .

$pH=−log(0.01885)=1.72_$

Hence, the $pH$ of a solution is $1.72_$ .

Conclusion

The $pH$ of a $Fe(NO3)3$ solution is $1.72_$ .

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Answer 4

Question

Chapter 17, Problem 17.51QP

Interpretation Introduction

Interpretation: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is to be calculated.

Concept introduction: The potential of $H+$ ions is called as $pH$ . The value of $pH$ is less than $7$ for acidic solutions and greater than $7$ for basic solutions. For neutral solution the value of $pH$ is $7$ .

To determine: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ .

Expert Solution & Answer

Answer to Problem 17.51QP

The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is $1.72_$ .

Explanation of Solution

The concentration of $Fe(NO3)3$ is $0.100 M$ .

The chemical equation for the dissociation of $Fe(NO3)3$ in water is,

$Fe(NO3)3+3H2O→Fe(OH)3+3HNO3$

The net ionic equation is,

$Fe3++3H2O→Fe(OH)3+3H+$

The $Ka$ of $Fe(NO3)3$ is $3.00×10−3$ .

The ICE table for the dissociation of $Fe(NO3)3$ is,

$Fe3+(aq)→Fe(OH)3(aq)+H+(aq)I0.10000C−xxxE0.100−xxx$

The concentrations of both $Fe(OH)3$ and $H+$ are equal and assumed to be $x$ and is calculated by the equilibrium constant expression,

$Ka=[Fe(OH)3][H+][Fe3+]$

Substitute the values of dissociation constant and concentration of base in the above expression to calculate the concentration of $H+$ ions.

$3.00×10−3=x20.100−xx2=3.00×10−3(0.100−x)x2=3.00×10−4−3.00×10−3x0=x2+3.00×10−3x−3.00×10−4$

Simplify the above equation by the formula,

$x=−b±b2−4ac2a$

Substitute the values of $a$ and $b$ in the above formula to calculate the value of $x$ .

$x=−(3.00×10−3)±(3.00×10−3)2−4×1×(−3.00×10−4)2×1=−3.00×10−3±9.0×10−6+1.2×10−32=−3.00×10−3+0.03472=0.01885 M$

Hence, the concentration of $H+$ is $0.01885 M$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above formula to calculate the value of $pH$ .

$pH=−log(0.01885)=1.72_$

Hence, the $pH$ of a solution is $1.72_$ .

Conclusion

The $pH$ of a $Fe(NO3)3$ solution is $1.72_$ .

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Answer 5

Chapter 17, Problem 17.51QP

Interpretation Introduction

Interpretation: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is to be calculated.

Concept introduction: The potential of $H+$ ions is called as $pH$ . The value of $pH$ is less than $7$ for acidic solutions and greater than $7$ for basic solutions. For neutral solution the value of $pH$ is $7$ .

To determine: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ .

Expert Solution & Answer

Answer to Problem 17.51QP

The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is $1.72_$ .

Explanation of Solution

The concentration of $Fe(NO3)3$ is $0.100 M$ .

The chemical equation for the dissociation of $Fe(NO3)3$ in water is,

$Fe(NO3)3+3H2O→Fe(OH)3+3HNO3$

The net ionic equation is,

$Fe3++3H2O→Fe(OH)3+3H+$

The $Ka$ of $Fe(NO3)3$ is $3.00×10−3$ .

The ICE table for the dissociation of $Fe(NO3)3$ is,

$Fe3+(aq)→Fe(OH)3(aq)+H+(aq)I0.10000C−xxxE0.100−xxx$

The concentrations of both $Fe(OH)3$ and $H+$ are equal and assumed to be $x$ and is calculated by the equilibrium constant expression,

$Ka=[Fe(OH)3][H+][Fe3+]$

Substitute the values of dissociation constant and concentration of base in the above expression to calculate the concentration of $H+$ ions.

$3.00×10−3=x20.100−xx2=3.00×10−3(0.100−x)x2=3.00×10−4−3.00×10−3x0=x2+3.00×10−3x−3.00×10−4$

Simplify the above equation by the formula,

$x=−b±b2−4ac2a$

Substitute the values of $a$ and $b$ in the above formula to calculate the value of $x$ .

$x=−(3.00×10−3)±(3.00×10−3)2−4×1×(−3.00×10−4)2×1=−3.00×10−3±9.0×10−6+1.2×10−32=−3.00×10−3+0.03472=0.01885 M$

Hence, the concentration of $H+$ is $0.01885 M$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above formula to calculate the value of $pH$ .

$pH=−log(0.01885)=1.72_$

Hence, the $pH$ of a solution is $1.72_$ .

Conclusion

The $pH$ of a $Fe(NO3)3$ solution is $1.72_$ .

Answer 6

Chapter 17, Problem 17.51QP

Interpretation Introduction

Interpretation: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is to be calculated.

Concept introduction: The potential of $H+$ ions is called as $pH$ . The value of $pH$ is less than $7$ for acidic solutions and greater than $7$ for basic solutions. For neutral solution the value of $pH$ is $7$ .

To determine: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ .

Expert Solution & Answer

Answer to Problem 17.51QP

The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is $1.72_$ .

Explanation of Solution

The concentration of $Fe(NO3)3$ is $0.100 M$ .

The chemical equation for the dissociation of $Fe(NO3)3$ in water is,

$Fe(NO3)3+3H2O→Fe(OH)3+3HNO3$

The net ionic equation is,

$Fe3++3H2O→Fe(OH)3+3H+$

The $Ka$ of $Fe(NO3)3$ is $3.00×10−3$ .

The ICE table for the dissociation of $Fe(NO3)3$ is,

$Fe3+(aq)→Fe(OH)3(aq)+H+(aq)I0.10000C−xxxE0.100−xxx$

The concentrations of both $Fe(OH)3$ and $H+$ are equal and assumed to be $x$ and is calculated by the equilibrium constant expression,

$Ka=[Fe(OH)3][H+][Fe3+]$

Substitute the values of dissociation constant and concentration of base in the above expression to calculate the concentration of $H+$ ions.

$3.00×10−3=x20.100−xx2=3.00×10−3(0.100−x)x2=3.00×10−4−3.00×10−3x0=x2+3.00×10−3x−3.00×10−4$

Simplify the above equation by the formula,

$x=−b±b2−4ac2a$

Substitute the values of $a$ and $b$ in the above formula to calculate the value of $x$ .

$x=−(3.00×10−3)±(3.00×10−3)2−4×1×(−3.00×10−4)2×1=−3.00×10−3±9.0×10−6+1.2×10−32=−3.00×10−3+0.03472=0.01885 M$

Hence, the concentration of $H+$ is $0.01885 M$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above formula to calculate the value of $pH$ .

$pH=−log(0.01885)=1.72_$

Hence, the $pH$ of a solution is $1.72_$ .

Conclusion

The $pH$ of a $Fe(NO3)3$ solution is $1.72_$ .

Answer 7

Chapter 17, Problem 17.51QP

Interpretation Introduction

Interpretation: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is to be calculated.

Concept introduction: The potential of $H+$ ions is called as $pH$ . The value of $pH$ is less than $7$ for acidic solutions and greater than $7$ for basic solutions. For neutral solution the value of $pH$ is $7$ .

To determine: The value of $pH$ for $0.100 M$ $Fe(NO3)3$ .

Answer 8

Expert Solution & Answer

Answer to Problem 17.51QP

The value of $pH$ for $0.100 M$ $Fe(NO3)3$ is $1.72_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The concentration of $Fe(NO3)3$ is $0.100 M$ .

The chemical equation for the dissociation of $Fe(NO3)3$ in water is,

$Fe(NO3)3+3H2O→Fe(OH)3+3HNO3$

The net ionic equation is,

$Fe3++3H2O→Fe(OH)3+3H+$

The $Ka$ of $Fe(NO3)3$ is $3.00×10−3$ .

The ICE table for the dissociation of $Fe(NO3)3$ is,

$Fe3+(aq)→Fe(OH)3(aq)+H+(aq)I0.10000C−xxxE0.100−xxx$

The concentrations of both $Fe(OH)3$ and $H+$ are equal and assumed to be $x$ and is calculated by the equilibrium constant expression,

$Ka=[Fe(OH)3][H+][Fe3+]$

Substitute the values of dissociation constant and concentration of base in the above expression to calculate the concentration of $H+$ ions.

$3.00×10−3=x20.100−xx2=3.00×10−3(0.100−x)x2=3.00×10−4−3.00×10−3x0=x2+3.00×10−3x−3.00×10−4$

Simplify the above equation by the formula,

$x=−b±b2−4ac2a$

Substitute the values of $a$ and $b$ in the above formula to calculate the value of $x$ .

$x=−(3.00×10−3)±(3.00×10−3)2−4×1×(−3.00×10−4)2×1=−3.00×10−3±9.0×10−6+1.2×10−32=−3.00×10−3+0.03472=0.01885 M$

Hence, the concentration of $H+$ is $0.01885 M$ .

The $pH$ of a solution is calculated by the formula,

$pH=−log[H+]$

Substitute the value of $[H+]$ in the above formula to calculate the value of $pH$ .

$pH=−log(0.01885)=1.72_$

Hence, the $pH$ of a solution is $1.72_$ .

Answer 12

Conclusion

The $pH$ of a $Fe(NO3)3$ solution is $1.72_$ .

Answer 13

Ch. 17.1 - Prob. 1PE Ch. 17.3 - Prob. 2PE Ch. 17.4 - Prob. 3PE Ch. 17.6 - Prob. 4PE Ch. 17.9 - Prob. 5PE Ch. 17.10 - Prob. 6PE Ch. 17 - Prob. 17.1VP Ch. 17 - Prob. 17.2VP Ch. 17 - Prob. 17.3VP Ch. 17 - Prob. 17.4VP

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