A 50.0-mL of 0.25 M HNO 2 is added to 50.0mL of 0.25M NaNO 2 . What is the conce

Question

A 50.0-mL of 0.25 M HNO2 is added to 50.0mL of 0.25M NaNO2.

What is the concentration of HNO2 and NaNO2 once two solutions are mixed?

What is the pH of this solution?   

20.0 mL of 0.15 M NaOH have been added to solution (b). Write a balanced equations showing what happens when NaOH is added to solution (b).

What is the CHANGE in pH when 20.0 mL of 0.15 M NaOH have been added to solution (b)?

Look over your answers to all the questions above as you do this one!   What is the CHANGE in pH of the solution when 20.0 mL of 0.15 M HCl have been added to solution (b)?

Explanation / Answer

for dilution

Use

M1V1 = M2V2

the final volume = 50 + 50 = 100 ml

first consider HN02

M1V1 = M2V2

0.25 x 50 = M2 x 100

M2 = 0.125

so the final conc of HN02 is 0.125 M

now consider NaN02

M1V1 = M2V2

0.25 x 50 = M2 x 100

M2 = 0.125

so the final conc of NaN02 is 0.125 M

we know that

Hn02 is a weak acid

now

HN02 and NaN02 forms a combination of weak acid and salt of weak acid

So they form a acidic buffer

for acidic buffers

pH = pKa + log [ salt / acid]

so

pH = pKa + log [ NaN02 / HN02]

also

pKa of HN02 is 3.398

so

pH = 3.398 + log [ 0.125 / 0.125]

pH = 3.398 + 0

pH = 3.398

so the pH of the final soltuion is 3.398

2) now 20 ml of 0.15 M NaOH is added

we know that

moles= molarity x volume (L)

so

moles of NaOH added = 0.15 x 20 x 10-3 = 3 x 10-3

moles of NaN02 = 0.125 x 100 x 10-3 = 12.5 x 10-3

moles of Hn02 = 0.125 x 100 x 10-3 = 12.5 x 10-3

the balanced reaction is

NaOH + HN02 ---> NaN02 + H20

from the above reaction

moles of HN02 reacted = moles of NaOH added = 3 x 10-3

moles of Hn02 remaining = 12.5 x 10-3 - 3 x 10-3 = 9.5 x 10-3

moles of NaN02 formed = moles of Hn02 reacted = 3 x 10-3

new moles of NaN02 = 12.5 x 10-3 + 3 x 10-3 = 15.5 x 10-3

now

pH = pKa + log [ NaN02 / HN02]

we know tha t

molarity = moles / volume(L)

as the final volume is same for both , they cancel out

so

ratio of molarities = ratio of moles

so

pH = 3.398 + log [ 15.5 x 10-3 / 9.5 x 10-3 ]

pH = 3.61

so the change in pH = final - initial

change in pH = 3.61 - 3.398

change in pH = 0.212

so the pH is increased by 0.212

3) now 20 ml of 0.15 HCl is added

we know that

moles= molarity x volume (L)

so

moles of HCl added = 0.15 x 20 x 10-3 = 3 x 10-3

moles of NaN02 = 0.125 x 100 x 10-3 = 12.5 x 10-3

moles of Hn02 = 0.125 x 100 x 10-3 = 12.5 x 10-3

the balanced reaction is

HCl + NaN02 ---> HN02 + NaCl

from the above reaction

moles of NaN02 reacted = moles of HCl added = 3 x 10-3

moles of Nan02 remaining = 12.5 x 10-3 - 3 x 10-3 = 9.5 x 10-3

moles of HN02 formed = moles of Nan02 reacted = 3 x 10-3

new moles of HN02 = 12.5 x 10-3 + 3 x 10-3 = 15.5 x 10-3

now

pH = pKa + log [ NaN02 / HN02]

we know tha t

molarity = moles / volume(L)

as the final volume is same for both , they cancel out

so

ratio of molarities = ratio of moles

so

pH = 3.398 + log [ 9.5 x 10-3 / 15.5 x 10-3 ]

pH = 3.186

so the change in pH = final - initial

change in pH = 3.186 - 3.398

change in pH = -0.212

so the pH is decreased by 0.212

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