a) Explanation The P H of the buffer solution can be obtained from Henderson-Hasselbalch equation. For H N O 2 and N a N O 2 buffer system the equation is P H = P K a + log [ N a N O 2 ] [ H N O 2 ] On plugging the given values in the equation P H = 3.15 + log 0.30 0.30 Thus P H = 3.15 . So P H of the buffer solution is 3.15. Now to change the P H 1 unit i.e. ( 3.15 ± 1 ), we need strong base to increase and strong acid to decrease. Let assume x mole of strong base MOH is added to the buffer solution and the P H become (3.15+1) = 4.15. The molarity of x mole of strong base MOH in the buffer solution will be x 100 × 1000 = 10 x ( M ) . Now as per the buffer mechanism the added base will fully dissociate in the solution in the solution and will produce 10 x ( M ) O H − ion which will react with H + ion to form water and the dissociation equilibrium of H N O 2 will be disturbed as per La-Chatelier’s principle b) Interpretation Introduction To determine: Buffer capacity of the solution of H N O 2 - N a N O 2 system and to compare with the other set.

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Author: Robinson

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Chapter 17, Problem 17.65SP

Interpretation Introduction

To determine:

Buffer capacity of the solution of HNO2 - NaNO2 system and to compare with the other set.

Interpretation Introduction

To determine:

Buffer capacity of the solution of HNO2 - NaNO2 system and to compare with the other set.

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Chapter 17, Problem 17.64SP

Chapter 17, Problem 17.66SP

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Chapter 17 Solutions

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Buffer capacity of the solution of H N O 2 - N a N O 2 system and to compare with the other set.

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To determine: Buffer capacity of the solution of HNO2 - NaNO2 system and to compare with the other set.

To determine: Buffer capacity of the solution of HNO2 - NaNO2 system and to compare with the other set.

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Chapter 17 Solutions

To determine:

Buffer capacity of the solution of HNO2 - NaNO2 system and to compare with the other set.

To determine:

Buffer capacity of the solution of HNO2 - NaNO2 system and to compare with the other set.