Calculating pH of 0.140M NaOCN Solution - Ka HOCN=3.5x10^-4

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SUMMARY

The pH of a 0.140M sodium cyanate (NaOCN) solution can be calculated using the dissociation constant (Ka) of the weak acid, HOCN, which is 3.5x10^-4. The relationship between Ka and Kb can be established using the formula Kb = Kw / Ka, where Kw is the ion product of water (1.0x10^-14 at 25°C). By calculating Kb, one can determine the hydroxide ion concentration (OH-) and subsequently the pOH, which can be converted to pH using the formula pH = 14 - pOH.

PREREQUISITES
  • Understanding of acid-base chemistry and dissociation constants
  • Familiarity with the concept of conjugate bases
  • Knowledge of the ion product of water (Kw)
  • Ability to perform logarithmic calculations for pH and pOH
NEXT STEPS
  • Calculate Kb for HOCN using the formula Kb = Kw / Ka
  • Determine the hydroxide ion concentration (OH-) from Kb and the concentration of NaOCN
  • Calculate pOH from the hydroxide ion concentration
  • Convert pOH to pH using the formula pH = 14 - pOH
USEFUL FOR

Chemistry students, educators, and professionals involved in analytical chemistry or environmental science, particularly those focusing on acid-base equilibria and solution chemistry.

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what is the pH of a 0.140M NaOCN solution? ka HOCN=3.5x10^-4
(rule of 100 can simplify 0.140-x to 0.140)

i know that ka=HOCN*OH^-/OCN^-=x^2/0.140
but i don't know the left hand side...i don't know what to do with HOCN=3.5x10^-4..
 
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OCN- is a weak, conjugate base of HOCN. Calculate Kb, calculate pOH from known Kb and concentration, convert to pH.

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