1) Calculate how many moles of KOOCH would be needed to create 1.00 L of a pH =

Question

1) Calculate how many moles of KOOCH would be needed to create 1.00 L of a pH = 9.069 solution. Assume the volume does not change as the solution is formed.

2) A 0.029 mol sample of a weak acid, HA, is dissolved in 541 mL of water and titrated with 0.41 M NaOH. After 31 mL of the NaOH solution has been added, the overall pH = 4.953. Calculate the Ka value for HA.

3) Consider a the titration of 0.795 L of 0.677 M carbonic acid (H2CO3) with 1.57 M NaOH. What is the pH at the second equivalence point of the titration?

Explanation / Answer

1) Calculate how many moles of KOOCH would be needed to create 1.00 L of a pH = 9.069 solution. Assume the volume does not change as the solution is formed.

HCOO- + H2O <==> HCOOH + OH-

since pH=-log[H+]=9.069

[H+]=antilog(-9.069)=8.53E-10

[OH-]=kw/[H+]=1.0E-14/8.53E-10=1.17E-5M

[HCOOH]=0.0000117M

pka=3.77

pkb=10.23

pOH=pkb+log[HCOOH]/[HCOO-]

14-9.069=10.23+log(0.0000117)-log[HCOOH]

log[HCOOH]=5.298-4.931=0.367

[HCOOH]=antilog(0.367)=2.328M

now [KCOOH]=[HCOOH]+[HCOO-]=2.328+0.0000117=2.328Moles

2) A 0.029 mol sample of a weak acid, HA, is dissolved in 541 mL of water and titrated with 0.41 M NaOH. After 31 mL of the NaOH solution has been added, the overall pH = 4.953. Calculate the Ka value for HA. (For the retake enter your answer in the following format = 1.2e3 using only 1 decimal place.)

3) Consider a the titration of 0.795 L of 0.677 M carbonic acid (H2CO3) with 1.57 M NaOH. What is the pH at the second equivalence point of the titration?

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