# 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?

• There are some calculations in this answer chemistry.stackexchange.com/questions/60068/… – porphyrin May 21 '19 at 18:54
• 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. – Poutnik May 21 '19 at 20:36

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

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