Calculate the pH of buffer solution

PROBLEM: A solution was prepared by mixing $\pu{25.00 mL}$ of $\ce{NaOH}$ $\pu{(c = 1mol/L)}$ and $\pu{10.00 mL}$ of acetic acid $\pu{(c = 2.5 mol/L)}$ in water to give $\pu{250 mL}$ of solution. 


What is the $\mathrm{pH}$?


$K\_\mathrm{a}=\pu{1.76\times 10^{-5} mol/L}$


Ok, so I know acetic acid will react with sodium hydroxide do form sodium acetate:


$$\ce{CH3COOH + NaOH -> CH3COONa + H2O}$$


initial quantities of $\ce{CH3COOH}$ and sodium hydroxide are $\pu{ 0.025 mol}$ (c\*V)each


after the equilibrium is reached all base and acid have reacted, so we have $\pu{0.025 mol }$ of $\ce{CH3COONa}$


I can't calculate the $\mathrm{pH}$ using Henderson-Hasselbalch approximation, because concentration of acid is 0.


I got a response that I should solve it like this (after I do ICE table like I did above):


$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$


$[\ce{OH-}]=\sqrt{ K\_\mathrm{b}\cdot c\_\mathrm{salt}}$
and after I calculate $\mathrm{pOH}=5.12$ I can calculate the $\mathrm{pH}=8.88$


I dont understand the last part, why is this so? what is 


$$\ce{CH3COO- + H2O <=> CH3COOH + OH- }$$


representing?


Thanks for answering! :)





> 
> I can't calculate the $\mathrm{pH}$ using Henderson-Hasselbalch approximation, because the concentration of acid is 0.
> 
> 
> 


Yes, as commented above, sodium acetate solution cannot be considered as buffer;the result at the equilibrium , will be the same as dissolving $\pu{0.025 mol }$ of $\ce{CH3COONa}$ in $\pu{250 ml}$ water,so you have the salt solution of $\pu{0.1M}\ce{[CH3COONa]}(c\_\mathrm{\text{salt}}=\ce{[CH3COONa]\_\mathrm{I}}=\pu{0.1M})$



> 
> I don't understand the last part, why is this so? what is...
> 
> 
> 


The salt ($\ce{ CH3COONa}$) is electrically neutral substance formed by cation $\ce{Na+}$ (an acid) and anion $\ce{ CH3COO−}$ (a base), completely dissociated or ionized in an aqueous solution as :
 $$\ce{CH3COONa -> Na^+\_\mathrm{(aq)} + CH3COO^−\_\mathrm{(aq)}}$$ 
a) Since $\ce{Na+}$ is the conjugate acid of a strong base, it won't be strong enough to react with water; $\ce{Na+}$ actually spectator ion.


b) Meanwhile, since $\ce{ CH3COO−}$ is the conjugate base of a weak acid, and therefore strong enough to be able to hydrolyze and accept ions $\ce{ H+}$ from water, so water act as an acid leaving a hydroxide ion $\ce{OH−}$ as :
$$\ce{CH3COO^− +H2O <=>CH3COOH + OH^−}$$ $$(K\_\mathrm{b(\ce{CH3COO-})}=\frac{K\_\mathrm{w}}{K\_\mathrm{a(\ce{CH3COOH})}}=\frac{10^{-14}}{1.76\times 10^{-5} }=5.68\times{10^{-10}})$$


c)The equilibrium equation of the hydrolysis of the conjugate base $\ce{CH3COO-}$:
$$ K\_\mathrm{b} =\frac{[\ce{OH-}]\_\text{equilibrium}[\ce{CH3COOH}]\_\text{equilibrium}}{[\ce{CH3COO-}]\_\text{equilibrium}}$$


Neglecting water autoionization and assuming :${\ce{[OH-]\_\text{equilibrium}}=\ce{[CH3COOH]\_\text{equilibrium}}}$, ${[\ce{CH3COO-}]\_\text{equilibrium}=c\_\mathrm{\text{salt}}=[\ce{CH3COONa}]\_\mathrm{I}}=\pu{0.1 M}$,so:
$$ K\_\mathrm{b} =\frac{[\ce{OH-}]\_\text{equilibrium}^2}{c\_\mathrm{\text{salt}}}$$


$$[\ce{OH-}]\_\text{equilibrium}=\sqrt{ K\_\mathrm{b}\cdot c\_\mathrm{salt}}$$
**You can calculate $\ce{[H+]}$** -without neglecting the water autoionization and without approximation- by substituting in the following cubic equation and solving by[wolframalpha](https://www.wolframalpha.com/input/?i=X%5E3%2B(1.76*10%5E-5%2B0.1)*X%5E2+-1*10%5E-14*X-1.76*10%5E-5*1*10%5E-14%3D0)


$$\
 [\ce{H+}]^3 
+( K\_\mathrm{a}+[\ce{CH3COONa}]\_0) [\ce{H+}]^2 
 -K\_\mathrm{w}[\ce{H+}] - K\_\mathrm{w} K\_\mathrm{a}
 $$
$\ce{[H+]} =1.32658 ,\pu{pH}=8.877$


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