A. Calculate the pH of a buffer solution prepared by adding 6.15 g of RbC 3 H 3

Question

A. Calculate the pH of a buffer solution prepared by adding 6.15 g of RbC3H3O2 to 250.0 mL of 0.165 M HC3H3O2. For HC3H3O2, Ka = 5.15x10-5

B. Write the net ionic reaction for the neutralization reaction that occurs when HBr(aq) is added to the buffer solution in part a.

C. Calculate the pH after adding 75.0mL of 0.195 M HBr (aq) to teh 250.0 mL buffer solution in part a.

D. Write the net ionic reaction for the neutralization reaction that occurs when NaOH (aq) is added to the buffer solution in part a.

E. Calculate the pH after adding 85.0 mL of 0.215 M NaOH to the 250.0 mL buffer solution in part a.

Explanation / Answer

A) given 6.15 grams of RbC3H302

moles of RbC3H302 = mass / molar mass

moles of RbC3H302 = 6.15 / 156.46

moles of RbC3H302 = 0.0393

moles of HC3H302 = molarity x volume (ml) /1000

moles of HC3H302 = 0.165 x 250 /1000

moles of HC3H302 = 0.04125

given is a acid buffer

using henderson hasselbach equation we get

pH = pKa + log [salt / acid]

pH = -logKa + log [RbC3H302 / HC3H302 ]

pH = -log 5.15 x 10-5 + log [ 0.0393 / 0.04125]

pH = 4.267

so pH is 4.267

B) when HBr is added , it reacts with RbC3H302

the reaction is given by

  C3H302- + HBr ------> HC3H302 + Br-

C) moles of HBr added = 0.195 x 75 /1000

moles of HBr added = 0.014625

from the above reaction is is clear that

0.014625 of HBr reacts with 0.014625 of RbC3H302 to give 0.014625 of HC3H302

new moles of RbC3H302 = 0.0393 - 0.014625

new moles of RbC3H302 = 0.024675

new moles of HC3H302 = 0.014625 + 0.04125

new moles of HC3H302 = 0.055875

so

pH = -logKa + log [RbC3H302 / HC3H302 ]

pH = -log 5.15 x 10-5 + log [ 0.024675  / 0.055875 ]

pH = 3.933

so the new pH is 3.933

D) when NaOH is added , it reacts with HC3H302

the reaction is given by

HC3H302 + NaOH ------> C3H302- + OH-

E) moles of NaOH added = 0.215 x 85 /1000

moles of NaOH added = 0.018275

from the above reaction is is clear that

0.018275  of NaOH reacts with 0.018275  of HC3H302 to give 0.018275 of C3H302-

new moles of RbC3H302 = 0.0393 + 0.018275

new moles of RbC3H302 = 0.057575

new moles of HC3H302 = 0.04125 - 0.018275

new moles of HC3H302 = 0.022975

so

pH = -logKa + log [RbC3H302 / HC3H302 ]

pH = -log 5.15 x 10-5 + log [ 0.057575 / 0.022975 ]

pH = 6.794

so the new pH is 6.794

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