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

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


Thanks for answering! :)

  • $\begingroup$ There are some calculations in this answer chemistry.stackexchange.com/questions/60068/… $\endgroup$
    – porphyrin
    May 21, 2019 at 18:54
  • $\begingroup$ Note that sodium acetate solution cannot be considered as buffer. It is very sensitive to excess of both sodium hydroxide and acetic acid, forming with the latter the acetic acid/sodium acetate buffer. $\endgroup$
    – Poutnik
    May 21, 2019 at 20:36

1 Answer 1


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 bywolframalpha

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


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