Can you anyone help me to set up a ice chart for this one?

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SUMMARY

The discussion focuses on setting up ICE (Initial, Change, Equilibrium) charts for calculating the initial concentrations of two bases, LiOH and HCN, given a pH of 9.72. For LiOH, it is established that an ICE chart is unnecessary due to its classification as a strong base, which fully dissociates. For HCN, participants clarify that the ICE chart can be constructed similarly to that for acids, emphasizing the need to calculate hydroxide ion concentration and its impact on pH.

PREREQUISITES
  • Understanding of pH and its calculation using the formula pH = -log[H+]
  • Knowledge of Brønsted-Lowry acid-base theory
  • Familiarity with ICE table construction for equilibrium reactions
  • Basic concepts of strong vs. weak bases
NEXT STEPS
  • Learn how to construct ICE tables for weak acids and bases
  • Study the dissociation constants (Ka and Kb) for weak acids and bases
  • Explore the relationship between pH, pOH, and ion concentrations
  • Investigate the properties and calculations involving strong bases like LiOH
USEFUL FOR

Chemistry students, educators, and anyone studying acid-base equilibria and the application of ICE charts in chemical calculations.

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Homework Statement



if the pH is 9.72 for the following 2 bases. what is the initial conc. of the bases
a)LiOH
b)HCN
the answer for a is 10^-4.28, but i don't know how to set up an ice chart

Homework Equations





The Attempt at a Solution

 
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For a), you don't need an ICE chart. You need pH=-log[H+] and [H+][OH-] = 10-14
 
chemisttree said:
For a), you don't need an ICE chart. You need pH=-log[H+] and [H+][OH-] = 10-14

Yes, but I still want to know what the ice chart like,
and for b, can you help me to set the ice chart, because HCN is weird
H3O+ + CN- <--> H2O + HCN
is this right ?
thanks!
 
CN- + H2O <-> HCN + OH-

Nothing weird, this is Brønsted-Lowry base.

Note: in the case of bases you have to calculate concentration of OH-, then to convert it to pH.

ICE tables are built exactly the same way as for acids, just OH- is the product (concentration of water in the case of cyanide is assumed to be not changing and ignored). You will be not able to solve LiOH with ICE table, as LiOH is considered to be a strong base, 100% dissociated, so final concentration of LiOH will be 0 - and you can't divide by zero.

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but,when i wrote out the the species
HCN(aq)
SA_____A
H30+ , H2O , CN-
_______B___ SB
Isn't the SA react with SB?
H3O+ + CN-?
 
Nope said:
HCN(aq)
SA_____A
H30+ , H2O , CN-
_______B___ SB

I am not familiar with this notation, please elaborate.

H3O+ + CN-?

These would react IF there would be enough H+ (H3O+). Concentration of water is much much larger, especially if you start with neutral solution.

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