(a) 9.11; (b) 9.26
<em>pH of original buffer
</em>
<em>Step 1</em>. Write the <em>chemical equation</em> for the equilibrium
Acid + H_2O ⇌ base + H_3O^(+)
; K_a = 5.69 × 10^(-10)
<em>Step 2</em>. Calculate the <em>pH of the buffer
</em>
The <em>Henderson-Hasselbalch equation</em> is pH = p<em>K</em>_a + log([base]/[acid]).
pH = -log[5.69 × 10^(-10)] + log[(0.170 mol)/L)/(0.187 mol/L)] = 9.24 + log0.1091
= 9.24 – 0.0414 = 9.20
<em>(a) pH after adding HCl
</em>
<em>Step 1</em>. Calculate the <em>moles of acid and base in the buffer</em>
Moles of acid = 200 mL × (0.187 mmol/1 mL) = 37.4 mmol
Moles of base = 200 mL × (0.170 mmol/1 mL) = 34.0 mmol
<em>Step 2</em>. Calculate the <em>moles of HCl added</em>
Moles of HCl added = 20.0 mL × (0.100 mmol/1 mL) = 2.00 mmol
<em>Step 3</em>. Calculate the <em>new moles of acid and base</em>
New moles of acid = (37.4 + 2.00) mmol = 39.4 mmol
New moles of base = (34.0 – 2.00) mmol = 32.0 mmol
<em>Step 4</em>. Calculate the <em>new pH
</em>
pH = 9.20 + log[(32.0 mmol)/(39.4 mmol)]) = 9.20 + log0.8122 = 9.20 - 0.0903 = 9.11
<em>(b) pH after adding NaOH
</em>
<em>Step 1</em>. Calculate the <em>moles of NaOH added</em>
Moles of NaOH added = 20.0 mL × (0.100 mmol/1 mL) = 2.00 mmol
<em>Step 2</em>. Calculate the <em>new moles of acid and base</em>
Moles of acid = (0.187 - 0.0200) = 0.167 mol
Moles of base = (0.170 + 0.0200) = 0.190 mol
<em>Step 3</em>. Calculate the <em>new pH
</em>
pH = 9.20 + log[(0.190 mol/L)/(0.167 mol/L)]) = 9.20 + log1.138 = 9.20 + 0.0560 = 9.26