(a) 4.75; (b) 4.81
<h2>(a) pH of buffer</h2>
<em>Step 1.</em> Write the equation for the equilibrium
HC2H3O2 + H2O ⇌ C2H3O2^(-) + H3O^(+); pKa = 4.75
or
HA + H2O ⇌ A^(-) + H3O^(+)
<em>Step 2.</em> Calculate the pH of the buffer
The Henderson-Hasselbalch equation is
pH = pKa + log([A^(-)]/[HA]) .
pH = 4.75 + log((0.300 mol/L)/(0.300 mol/L)) = 4.75 + log1.00 = 4.75 + 0
= 4 .75
<h2 /><h2>(b) pH of buffer after adding NaOH</h2>
<em>Step 1.</em> Calculate the new concentrations of HA and A^(-)
We can use an ICE table for the calculation.
_______HA + H2O ⇌ H3O^(+) + A^(-)
I/mol:__0.300_______0.300
C/mol:_-0.020______+0.020
E/mol:__0.280_______0.320
pH = pKa + log([A^(-)]/[HA]) = 4.75 + log((0.320 mol/L)/(0.280 mol/L))
= 4.75 + log1.143 = 4.75 + 0.058 = 4 .81
This result makes sense because adding a base to a buffer should increase its pH.