A buffer solution contains 0.255 M K HCO 3 and 0.440 M K 2 CO 3 . If 0.0348 mole

Question

A buffer solution contains 0.255 M KHCO3 and 0.440 M K2CO3.

If 0.0348 moles of nitric acid are added to 250. mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume does not change upon adding nitric acid)

pH =

-------------------------

A buffer solution contains 0.412 M ammonium bromide and 0.360 M ammonia.

If 0.0578 moles of sodium hydroxide are added to 250 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume does not change upon adding sodium hydroxide)

pH =

----------------------

A buffer solution contains 0.232 M hydrofluoric acid and 0.436 M sodium fluoride.

If 0.0290 moles of hydroiodic acid are added to 125 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume does not change upon adding hydroiodic acid)

pH =

Explanation / Answer

1)

we know that

moles = molarity x volume (ml) /1000

so

moles of KHC03 = 0.255 x 250 / 1000 = 0.06375

moles of K2C03 = 0.44 x 250 / 1000 = 0.11

now

moles of HN03 added = 0.0348

we know that

HN03 is a acid. So it reacts with base K2C03

so

C032- + H+ ---> HC03-

we can see that

moles of K2C03 reacted = moles of H+ added = 0.0348

moles of HC03- formed = moles of H+ added = 0.0348

so

finally

moles of K2C03 = 0.11 - 0.0348 = 0.0752

moles of KHC03 = 0.06375 + 0.0348 = 0.09855

now

for buffers

pH = pKa+ log [conjugate base /acid]

pH = pKa + log [ K2C03 / KHC03]

pH = 10.32 + log [ 0.0752 / 0.09855]

pH = 10.2

so

pH of the solution is 10.2 after adding nitric acid

2)

we know that

moles = molarity x volume (ml) /1000

so

moles of NH4Br = 0.412 x 250 / 1000 = 0.103

moles of NH3 = 0.36 x 250 / 1000 = 0.09

now

moles of NaOH added = 0.0578

we know that

NaOH is a base. So it reacts with base NH4Br

so

NH4+ + OH- ---> NH3 + H20

we can see that

moles of NH4+ reacted = moles of OH- added = 0.0578

moles of NH3 formed = moles of H+ added = 0.0578

so

finally

moles of NH4Br = 0.103 - 0.0578 = 0.0452

moles of NH3 = 0.09 + 0.0578 = 0.1478

now

for buffers

pOH = pKb + log [conjugate acid /base]

pOH = pKb + log [ NH4Br / NH3]

pOH = 4.75 + log [ 0.0452 / 0.1478]

pOH = 4.235

now

pH = 14 - pOH

pH = 14 - 4.235

pH = 9.765

so

pH of the solution is 9.765 after adding NaOH

3)

we know that

moles = molarity x volume (ml) /1000

so

moles of HF = 0.232 x 125 / 1000 = 0.029

moles of NaF = 0.436 x 125 / 1000 = 0.0545

now

moles of HI added = 0.029

we know that

HI is a acid. So it reacts with base NaF

so

F- + H+ ---> HF

we can see that

moles of NaF reacted = moles of H+ added = 0.029

moles of HF formed = moles of H+ added = 0.029

so

finally

moles of NaF = 0.0545 - 0.029 = 0.0255

moles of HF = 0.029 + 0.029 = 0.058

now

for buffers

pH = pKa + log [conjugate base /acid]

pH = pKa + log [ NaF / HF]

pH = 3.17 + log [ 0.0255 / 0.058]

pH = 2.813

so

pH of the solution is 2.813 after adding hydroiodic acid

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