The concentration of sodium hydroxide has to be determined using the given data. Concept introduction: Molarity is defined as strength of substance in terms of amount of moles of solute dissolved in one litre of solution. The molarity of HNO 2 and NaOH

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Answer 1

Question

Chapter 17, Problem 17.72QP

Interpretation Introduction

Interpretation:

The concentration of sodium hydroxide has to be determined using the given data.

Concept introduction:

Molarity is defined as strength of substance in terms of amount of moles of solute dissolved in one litre of solution.

The molarity of $HNO2$ and $NaOH$

Expert Solution & Answer

Answer to Problem 17.72QP

The concentration of sodium hydroxide is $1.28M$

Explanation of Solution

Given,

$Strength of HNO2 is 2.00M$

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 2.00 0 0 Change in concentration(M): -x +x +x Equilibrium concentration(M): 2.00-x x x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = x22.00 - x ≈ x22.00 x = [H+] = 0.030M pH = -log(0.030) = 1.52$

The $pH$ after addition of sodium hydroxide is $1.5$ times greater

$pH=1.52+1.50=3.02 [H+]= 10-pH = 10-3.02 = 9.55 × 10-4MAfter addition of NaOH, 2.00M HNO2 get diluted MiVi = MfVf (2.00M)(400 mL) = Mf(600 mL) Mf = 1.33M$

The $pH$ of nitrous acid is calculated using acid dissociation constant ( $Ka$ ). From the $pH$ , the molarity of nitrous acid after addition of sodium hydroxide is calculated.

Now we have buffer solution and not reached the equivalence point.

$HNO2(aq)+NaOH(aq) → NaNO2(aq)+H2O(l)$

Because the nitrous acid and sodium hydroxide are in same ratio, the reduction in concentration of nitrous acid is same as sodium hydroxide.

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 1.33 0 0 Change in concentration(M): -x +x Equilibrium concentration(M): 1.33-x 9.55×10-4 x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = (9.55 × 10-4)(x)1.33 - x x = 0.426M$

After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ has to calculate for $600mL$

$MiVi = MfVfMi(200 mL) = (0.426M)(600 mL) [NaOH] = Mi = 1.28M$

The concentration of nitrous acid is calculated using acid dissociation constant expression. After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ is calculated for $600mL$

Conclusion

The concentration of sodium hydroxide was determined using the given data.

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Answer 2

Question

Chapter 17, Problem 17.72QP

Interpretation Introduction

Interpretation:

The concentration of sodium hydroxide has to be determined using the given data.

Concept introduction:

Molarity is defined as strength of substance in terms of amount of moles of solute dissolved in one litre of solution.

The molarity of $HNO2$ and $NaOH$

Expert Solution & Answer

Answer to Problem 17.72QP

The concentration of sodium hydroxide is $1.28M$

Explanation of Solution

Given,

$Strength of HNO2 is 2.00M$

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 2.00 0 0 Change in concentration(M): -x +x +x Equilibrium concentration(M): 2.00-x x x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = x22.00 - x ≈ x22.00 x = [H+] = 0.030M pH = -log(0.030) = 1.52$

The $pH$ after addition of sodium hydroxide is $1.5$ times greater

$pH=1.52+1.50=3.02 [H+]= 10-pH = 10-3.02 = 9.55 × 10-4MAfter addition of NaOH, 2.00M HNO2 get diluted MiVi = MfVf (2.00M)(400 mL) = Mf(600 mL) Mf = 1.33M$

The $pH$ of nitrous acid is calculated using acid dissociation constant ( $Ka$ ). From the $pH$ , the molarity of nitrous acid after addition of sodium hydroxide is calculated.

Now we have buffer solution and not reached the equivalence point.

$HNO2(aq)+NaOH(aq) → NaNO2(aq)+H2O(l)$

Because the nitrous acid and sodium hydroxide are in same ratio, the reduction in concentration of nitrous acid is same as sodium hydroxide.

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 1.33 0 0 Change in concentration(M): -x +x Equilibrium concentration(M): 1.33-x 9.55×10-4 x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = (9.55 × 10-4)(x)1.33 - x x = 0.426M$

After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ has to calculate for $600mL$

$MiVi = MfVfMi(200 mL) = (0.426M)(600 mL) [NaOH] = Mi = 1.28M$

The concentration of nitrous acid is calculated using acid dissociation constant expression. After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ is calculated for $600mL$

Conclusion

The concentration of sodium hydroxide was determined using the given data.

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Answer 3

Question

Chapter 17, Problem 17.72QP

Interpretation Introduction

Interpretation:

The concentration of sodium hydroxide has to be determined using the given data.

Concept introduction:

Molarity is defined as strength of substance in terms of amount of moles of solute dissolved in one litre of solution.

The molarity of $HNO2$ and $NaOH$

Expert Solution & Answer

Answer to Problem 17.72QP

The concentration of sodium hydroxide is $1.28M$

Explanation of Solution

Given,

$Strength of HNO2 is 2.00M$

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 2.00 0 0 Change in concentration(M): -x +x +x Equilibrium concentration(M): 2.00-x x x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = x22.00 - x ≈ x22.00 x = [H+] = 0.030M pH = -log(0.030) = 1.52$

The $pH$ after addition of sodium hydroxide is $1.5$ times greater

$pH=1.52+1.50=3.02 [H+]= 10-pH = 10-3.02 = 9.55 × 10-4MAfter addition of NaOH, 2.00M HNO2 get diluted MiVi = MfVf (2.00M)(400 mL) = Mf(600 mL) Mf = 1.33M$

The $pH$ of nitrous acid is calculated using acid dissociation constant ( $Ka$ ). From the $pH$ , the molarity of nitrous acid after addition of sodium hydroxide is calculated.

Now we have buffer solution and not reached the equivalence point.

$HNO2(aq)+NaOH(aq) → NaNO2(aq)+H2O(l)$

Because the nitrous acid and sodium hydroxide are in same ratio, the reduction in concentration of nitrous acid is same as sodium hydroxide.

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 1.33 0 0 Change in concentration(M): -x +x Equilibrium concentration(M): 1.33-x 9.55×10-4 x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = (9.55 × 10-4)(x)1.33 - x x = 0.426M$

After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ has to calculate for $600mL$

$MiVi = MfVfMi(200 mL) = (0.426M)(600 mL) [NaOH] = Mi = 1.28M$

The concentration of nitrous acid is calculated using acid dissociation constant expression. After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ is calculated for $600mL$

Conclusion

The concentration of sodium hydroxide was determined using the given data.

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Answer 4

Question

Chapter 17, Problem 17.72QP

Interpretation Introduction

Interpretation:

The concentration of sodium hydroxide has to be determined using the given data.

Concept introduction:

Molarity is defined as strength of substance in terms of amount of moles of solute dissolved in one litre of solution.

The molarity of $HNO2$ and $NaOH$

Expert Solution & Answer

Answer to Problem 17.72QP

The concentration of sodium hydroxide is $1.28M$

Explanation of Solution

Given,

$Strength of HNO2 is 2.00M$

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 2.00 0 0 Change in concentration(M): -x +x +x Equilibrium concentration(M): 2.00-x x x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = x22.00 - x ≈ x22.00 x = [H+] = 0.030M pH = -log(0.030) = 1.52$

The $pH$ after addition of sodium hydroxide is $1.5$ times greater

$pH=1.52+1.50=3.02 [H+]= 10-pH = 10-3.02 = 9.55 × 10-4MAfter addition of NaOH, 2.00M HNO2 get diluted MiVi = MfVf (2.00M)(400 mL) = Mf(600 mL) Mf = 1.33M$

The $pH$ of nitrous acid is calculated using acid dissociation constant ( $Ka$ ). From the $pH$ , the molarity of nitrous acid after addition of sodium hydroxide is calculated.

Now we have buffer solution and not reached the equivalence point.

$HNO2(aq)+NaOH(aq) → NaNO2(aq)+H2O(l)$

Because the nitrous acid and sodium hydroxide are in same ratio, the reduction in concentration of nitrous acid is same as sodium hydroxide.

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 1.33 0 0 Change in concentration(M): -x +x Equilibrium concentration(M): 1.33-x 9.55×10-4 x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = (9.55 × 10-4)(x)1.33 - x x = 0.426M$

After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ has to calculate for $600mL$

$MiVi = MfVfMi(200 mL) = (0.426M)(600 mL) [NaOH] = Mi = 1.28M$

The concentration of nitrous acid is calculated using acid dissociation constant expression. After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ is calculated for $600mL$

Conclusion

The concentration of sodium hydroxide was determined using the given data.

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Answer 5

Chapter 17, Problem 17.72QP

Interpretation Introduction

Interpretation:

The concentration of sodium hydroxide has to be determined using the given data.

Concept introduction:

Molarity is defined as strength of substance in terms of amount of moles of solute dissolved in one litre of solution.

The molarity of $HNO2$ and $NaOH$

Expert Solution & Answer

Answer to Problem 17.72QP

The concentration of sodium hydroxide is $1.28M$

Explanation of Solution

Given,

$Strength of HNO2 is 2.00M$

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 2.00 0 0 Change in concentration(M): -x +x +x Equilibrium concentration(M): 2.00-x x x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = x22.00 - x ≈ x22.00 x = [H+] = 0.030M pH = -log(0.030) = 1.52$

The $pH$ after addition of sodium hydroxide is $1.5$ times greater

$pH=1.52+1.50=3.02 [H+]= 10-pH = 10-3.02 = 9.55 × 10-4MAfter addition of NaOH, 2.00M HNO2 get diluted MiVi = MfVf (2.00M)(400 mL) = Mf(600 mL) Mf = 1.33M$

The $pH$ of nitrous acid is calculated using acid dissociation constant ( $Ka$ ). From the $pH$ , the molarity of nitrous acid after addition of sodium hydroxide is calculated.

Now we have buffer solution and not reached the equivalence point.

$HNO2(aq)+NaOH(aq) → NaNO2(aq)+H2O(l)$

Because the nitrous acid and sodium hydroxide are in same ratio, the reduction in concentration of nitrous acid is same as sodium hydroxide.

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 1.33 0 0 Change in concentration(M): -x +x Equilibrium concentration(M): 1.33-x 9.55×10-4 x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = (9.55 × 10-4)(x)1.33 - x x = 0.426M$

After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ has to calculate for $600mL$

$MiVi = MfVfMi(200 mL) = (0.426M)(600 mL) [NaOH] = Mi = 1.28M$

The concentration of nitrous acid is calculated using acid dissociation constant expression. After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ is calculated for $600mL$

Conclusion

The concentration of sodium hydroxide was determined using the given data.

Answer 6

Chapter 17, Problem 17.72QP

Interpretation Introduction

Interpretation:

The concentration of sodium hydroxide has to be determined using the given data.

Concept introduction:

Molarity is defined as strength of substance in terms of amount of moles of solute dissolved in one litre of solution.

The molarity of $HNO2$ and $NaOH$

Expert Solution & Answer

Answer to Problem 17.72QP

The concentration of sodium hydroxide is $1.28M$

Explanation of Solution

Given,

$Strength of HNO2 is 2.00M$

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 2.00 0 0 Change in concentration(M): -x +x +x Equilibrium concentration(M): 2.00-x x x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = x22.00 - x ≈ x22.00 x = [H+] = 0.030M pH = -log(0.030) = 1.52$

The $pH$ after addition of sodium hydroxide is $1.5$ times greater

$pH=1.52+1.50=3.02 [H+]= 10-pH = 10-3.02 = 9.55 × 10-4MAfter addition of NaOH, 2.00M HNO2 get diluted MiVi = MfVf (2.00M)(400 mL) = Mf(600 mL) Mf = 1.33M$

The $pH$ of nitrous acid is calculated using acid dissociation constant ( $Ka$ ). From the $pH$ , the molarity of nitrous acid after addition of sodium hydroxide is calculated.

Now we have buffer solution and not reached the equivalence point.

$HNO2(aq)+NaOH(aq) → NaNO2(aq)+H2O(l)$

Because the nitrous acid and sodium hydroxide are in same ratio, the reduction in concentration of nitrous acid is same as sodium hydroxide.

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 1.33 0 0 Change in concentration(M): -x +x Equilibrium concentration(M): 1.33-x 9.55×10-4 x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = (9.55 × 10-4)(x)1.33 - x x = 0.426M$

After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ has to calculate for $600mL$

$MiVi = MfVfMi(200 mL) = (0.426M)(600 mL) [NaOH] = Mi = 1.28M$

The concentration of nitrous acid is calculated using acid dissociation constant expression. After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ is calculated for $600mL$

Conclusion

The concentration of sodium hydroxide was determined using the given data.

Answer 7

Chapter 17, Problem 17.72QP

Interpretation Introduction

Interpretation:

The concentration of sodium hydroxide has to be determined using the given data.

Concept introduction:

Molarity is defined as strength of substance in terms of amount of moles of solute dissolved in one litre of solution.

The molarity of $HNO2$ and $NaOH$

Answer 8

Expert Solution & Answer

Answer to Problem 17.72QP

The concentration of sodium hydroxide is $1.28M$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given,

$Strength of HNO2 is 2.00M$

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 2.00 0 0 Change in concentration(M): -x +x +x Equilibrium concentration(M): 2.00-x x x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = x22.00 - x ≈ x22.00 x = [H+] = 0.030M pH = -log(0.030) = 1.52$

The $pH$ after addition of sodium hydroxide is $1.5$ times greater

$pH=1.52+1.50=3.02 [H+]= 10-pH = 10-3.02 = 9.55 × 10-4MAfter addition of NaOH, 2.00M HNO2 get diluted MiVi = MfVf (2.00M)(400 mL) = Mf(600 mL) Mf = 1.33M$

The $pH$ of nitrous acid is calculated using acid dissociation constant ( $Ka$ ). From the $pH$ , the molarity of nitrous acid after addition of sodium hydroxide is calculated.

Now we have buffer solution and not reached the equivalence point.

$HNO2(aq)+NaOH(aq) → NaNO2(aq)+H2O(l)$

Because the nitrous acid and sodium hydroxide are in same ratio, the reduction in concentration of nitrous acid is same as sodium hydroxide.

$HNO2(aq) → H+(aq) + NO2-(aq) Initial concentration(M): 1.33 0 0 Change in concentration(M): -x +x Equilibrium concentration(M): 1.33-x 9.55×10-4 x Ka = [H+][NO2-][HNO2] 4.5 × 10-4 = (9.55 × 10-4)(x)1.33 - x x = 0.426M$

After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ has to calculate for $600mL$

$MiVi = MfVfMi(200 mL) = (0.426M)(600 mL) [NaOH] = Mi = 1.28M$

The concentration of nitrous acid is calculated using acid dissociation constant expression. After dilution to $600mL$ the concentration of sodium hydroxide is same as decrease in concentration of nitrous acid. The concentration of $NaOH$ is calculated for $600mL$