1. Determine the pH of 100.0mL of a buffer solution containing 0.200M HNO2 and 0

Question

1. Determine the pH of 100.0mL of a buffer solution containing 0.200M HNO2 and 0.200M KNO2. Ka for HNO2 is 4.0x10-4

2. Calculate the pH if 30.00mL of 0.150M NaOH was added to the buffer from question 1. 3. Calculate the pH if 10.00mL of 2.00M HBr was added to the buffer from question 1. 4. Calculate the pH if 15.00mL of 2.00M HBr was added to the buffer from question 1.

3. Calculate the pH if 10.00mL of 2.00M HBr was added to the buffer from question 1.

4. Calculate the pH if 15.00mL of 2.00M HBr was added to the buffer from question 1.

Explanation / Answer

1)

we know that

pKa = -log Ka

pKa = -log 4 x 10-4

pKa = 3.398

now

we know that

for buffers

pH = pKa + log [salt / acid]

so in this case

pH = pKa + log [KN02 / HN02]

so

pH = 3.398 + log [0.2 / 0.2]

pH = 3.398

so

pH of the solution is 3.398

2)

we know that

moles = molarity x volume (ml) / 1000

so

initially

moles of HN02 = 0.2 x 100/ 1000 = 0.02

moles of KN02 = 0.2 x 100 / 1000 = 0.02

now

moles of NaOH added = 0.15 x 30 / 1000 = 0.0045

now

the reaction is

HN02 + OH- ---> N02- + H20

we can see that

moles of HN02 reacted = moles of NaOH added = 0.0045

moles of N02- formed = moles of NaOH added = 0.0045

so finally

moles of HN02 = 0.02 - 0.0045 = 0.0155

moles of N02- = 0.02 + 0.0045 = 0.0245

now

pH = pKa + log [N02- / HN02]

pH = 3.398 + log [ 0.0245 / 0.0155]

pH = 3.597

so

pH after addition of NaOH is 3.597

3)

now

moles of HBr added = 2 x 10 / 1000 = 0.02

now the reaction is

N02- + H+ --> HN02

we can see that

moles of N02- reacted = moles of HBr added = 0.02

moles of HN02 formed = moles of HBr added = 0.02

so finally

moles of N02- = 0.02 - 0.02 = 0

moles of HN02= 0.02 + 0.02 = 0.04

final volume = 100 + 10 = 110 ml

now

[HN02]= 0.04 x 1000 / 110

[HN02] = 0.36364

now

HN02 is a weak acid

for weak acids

[H+] = sqrt ( Ka x C)

so

[H+] = sqrt ( 4 x 10-4 x 0.36364)

[H+] = 0.012

now

pH = -log [H+]

pH = -log 0.012

pH = 1.92

so

pH of the solution after addition of HBr is 1.92

4) now

moles of HBr added = 2 x 15 / 1000 = 0.03

now the reaction is

N02- + H+ --> HN02

we can see that

moles of HBr reacted = moles of N02- added = 0.02

moles of HN02 formed = moles of N02- added = 0.02

so finally

moles of N02- = 0.02 - 0.02 = 0

moles of HN02= 0.02 + 0.02 = 0.04

moles of HBr = 0.03 - 0.02 = 0.01

final volume = 100 + 10 = 115 ml

now

[HBr] = 0.01 x 1000 / 115 = 0.087

[HN02] = 0.04 x 1000 / 115 = 0.348

now

for HN02

[H+] = sqrt ( 0.348 x 4 x 10-4) = 0.0118

now as HBr is a very strong acid

[H+] = [HBr] = 0.087

so

total [H+] = 0.0118 + 0.087 = 0.0988

now

pH = -log 0.0988

pH = 1.005

so

pH after addition of 15ml of HBr is 1.005

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