Calculating pH and POH: NaOH in Water Solution

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The discussion revolves around calculating the pH and hydronium ion concentration when 0.45g of NaOH is dissolved in 500ml of water. The concentration of hydroxyl ions ([OH-]) is determined to be 0.0225 M. To find the hydronium ion concentration ([H3O+]), the relationship between pH and pOH is utilized, specifically the equation pH + pOH = 14. The confusion arises regarding why 14 minus the pH value is used to find the hydronium concentration, which is explained by the water ion product equilibrium constant of 10^-14. Understanding this relationship is crucial for accurately calculating pH in strong base solutions.
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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
 
More datils: water ion product.
 
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