Calculating pH and POH: NaOH in Water Solution

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SUMMARY

The discussion focuses on calculating the pH and hydronium ion concentration of a solution containing 0.45g of NaOH mixed with 500ml of water. The participant correctly calculates the hydroxyl ion concentration as 0.0225 mol/L. To find the hydronium ion concentration, they apply the relationship between pH and pOH, utilizing the equation pH + pOH = 14. The participant seeks clarification on the rationale behind subtracting the pH from 14 to determine the hydronium ion concentration, referencing the water ion product equilibrium constant of 10-14.

PREREQUISITES
  • Understanding of strong bases, specifically NaOH
  • Knowledge of molar mass calculations
  • Familiarity with pH and pOH concepts
  • Basic principles of water ion product and equilibrium
NEXT STEPS
  • Study the relationship between pH, pOH, and the water ion product constant (Kw)
  • Learn how to calculate hydronium ion concentration from hydroxyl ion concentration
  • Explore the concept of autohydronolysis of water and its implications in acid-base chemistry
  • Practice problems involving strong bases and their effects on pH calculations
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Chemistry students, educators, and anyone interested in understanding acid-base equilibria and calculations involving strong bases like NaOH.

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Homework Statement



NaOH is an example of a strong base. if 0.45g of NaOH is mixed with 500ml of water what is the


[OH]
[H3O+]
pH



Homework Equations





The Attempt at a Solution



so, i calculating moles thru molar mass and divided by 0.5 to get mol/l. put that as the concentration of the hydroxyl. (0.0225)

What i don't understand is how to get the hydronium ion concentration. I know that its going to involve doing the -log of 0.025 to get the pH. Which is 1.4. But why do i do 14-1.4 then do the antilog of that to get the concentration of the hydronium ion?

I know how its solved, My real question is... why do i do 14- (ph of hydroxl) to find the concentration of the hydronium?
 
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Because water's autohydronolysis has the equilibrium constant of 10^-14
[OH-]*[H3O]=10^-14 <--> pOH- + pH = 14

WHen you pour a base into it, a negligible part of that base will be converted to water until the equilibrium condition is met
 

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