Calculate the pH of the solution that results from the addition of 0-040 moles of HNO 3 to a buffer made by combining 0-500 L of 0-380 M HC 3 H 5 O 2 K a -1-30 - 10-5 and 0-500 L of 0-380 M NaC 3 H 5 O 2- Assume addition of the nitric acid has no effect on volume-take log 1-30-0-114- log 0-652-0-186A- 7B- 5-65C- 3-24D- 4-7

The correct option is D 4.71) The nitric acid will reduce the amount of NaC_3H_5O_2: nbsp; 0.380 molL^ 1× 0.500 L = 0.190 mol nbsp; of NaC_3H_5O_2 0.19 ...

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Standard XII

Chemistry

Acidic Buffer Action

Calculate the...

Question

Calculate the pH of the solution that results from the addition of 0.040 moles of $H N O_{3}$ to a buffer made by combining 0.500 L of 0.380 M $H C_{3} H_{5} O_{2} (K_{a} = 1.30 \times 10^{- 5})$ and 0.500 L of 0.380 M $N a C_{3} H_{5} O_{2} .$ Assume addition of the nitric acid has no effect on volume.
$take l o g 1.30 = 0.114, l o g 0.652 = - 0.186$

4.7

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3.24

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Solution

The correct option is A 4.7
1) The nitric acid will reduce the amount of $N a C_{3} H_{5} O_{2} :$
$0.380 m o l L^{- 1} \times 0.500 L = 0.190 mol of N a C_{3} H_{5} O_{2}$
$0.190 m o l - 0.040 m o l = 0.150 m o l N a C_{3} H_{5} O_{2} remaining$

2) The reaction in 1 will increase the amount of $H C_{3} H_{5} O_{2}$ :
The increase will be by the same amount of the decrease
$0.190 m o l + 0.040 m o l = 0.230 m o l o f H C_{3} H_{5} O_{2}$
$p K_{a} = - l o g K_{a}$
$p K_{a} = - l o g 1.30 \times 10^{- 5})$
$p K_{a} = 5 - l o g 1.30$
$p K_{a} = 5 - 0.114 = 4.886$
The new $p H$ using Henderson-Hasselbalch equation:
$p H = p K_{a} + l o g ([salt] / [acid])$
$p H = p K_{a} + l o g ([N a C_{3} H_{5} O_{2}] / [H C_{3} H_{5} O_{2}])$
$p H = 4.886 + l o g (0.150 / 0.230) volume is same for both$
$p H = 4.886 + l o g (0.150 / 0.230)$
$p H = 4.886 + l o g (0.652)$
$p H = 4.886 - 0.186$
$p H = 4.70$

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Similar questions

Q. Calculate the pH of the solution that results from the addition of 0.040 moles of

H N O_{3}

to a buffer made by combining 0.500 L of 0.380 M

H C_{3} H_{5} O_{2} (K_{a} = 1.30 \times 10^{- 5})

and 0.500 L of 0.380 M

N a C_{3} H_{5} O_{2} .

Assume addition of the nitric acid has no effect on volume.

take l o g 1.30 = 0.114, l o g 0.652 = - 0.186

Q. Addition of sodium hydroxide solution to a weak acid

(H A)

results in a buffer of

p H = 6

. If ionisation constant of

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10^{- 5}

, the ratio of salt to acid concentration in the buffer solution will be:

Q. Calculate the change in pH of one litre of buffer solution containing 0.10 mole each of

N H_{3}

N H_{4} C l

upon addition of, (i) 0.02 mole of dissolved gaseous

H C l

(ii) 0.02 mole of dissolved

N a O H

. Assume no change in solution volume.

(K_{b} f o r N H_{3} = 1.8 \times 10^{- 5})

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{N H}_{3}

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Q. One litre of a buffer solution is prepared by dissolving

0.6

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{N H}_{3}

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{N H}_{4} C l

. What is the pH of the solution? for

{N H}_{3}

K_{b} = 1.85 \times 10^{- 5}

.
(i) What is the pH of the buffer after addition of

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(ii) What is the pH of the buffer after addition of

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Acidic Buffer Action

Standard XII Chemistry

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Calculate the pH of the solution that results from the addition of 0.040 moles of HNO3 to a buffer made by combining 0.500 L of 0.380 M HC3H5O2(Ka=1.30×10−5) and 0.500 L of 0.380 M NaC3H5O2. Assume addition of the nitric acid has no effect on volume. take log 1.30=0.114, log 0.652=−0.186