# Problem: Find the pH of the equivalence point(s) and the volume (mL) of 0.0372 M NaOH needed to reach the point(s) in titrations of(a) 42.2 mL of 0.0520 M CH 3COOH

###### FREE Expert Solution

NaOH → Na+ + OH

**Na+ → neutral ion

Reaction:      CH3COOH + OH  CH3COO + H2O

Step 1: Calculate the volume (mL) of 0.0372 M NaOH needed to reach the point(s)

molarity (volume) → moles

Recall:

$\overline{){\mathbf{M}}{\mathbf{=}}\frac{\mathbf{mol}}{\mathbf{L}}}$

*convert volumes from mL to L → 1 mL = 10-3 L

at the equivalence point of a titration:

$\overline{){\left(\mathbf{MV}\right)}_{{\mathbf{acid}}}{\mathbf{=}}{\left(\mathbf{MV}\right)}_{{\mathbf{base}}}}$

VNaOH58.99 mL

Step 2. Calculate the initial amounts of acid and base in moles before the reaction happens.

moles CH3COOH = 2.1944×10‒3 mol

moles CH3COOH = moles NaOH

moles NaOH = 2.1944×10‒3 mol

Step 3. Construct an ICF Chart.

After the reaction is complete the solution contains:

2.1944×10‒3 mol CH3COO  conjugate base

Step 4. Calculate the new concentration of the species left in the solution.

volume of solution = 0.10119 L

molarity CH3COO = 0.0217 M

Step 5. Calculate pH.

Since only the conjugate base is present after the titration and the conjugate base is a weak base, we will use an ICE chart to find pH.

We will need the Kb for the reaction.

Ka CH3COOH = 1.8×105

Ka and Kb are connected by the following equation:

$\overline{){{\mathbf{K}}}_{{\mathbf{w}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{·}}{{\mathbf{K}}}_{{\mathbf{b}}}}$

Where Kw is the autoionization constant of water:
Assuming the reaction is at 25°C:             Kw = 1.0x10-14

Calculate Kb:

$\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{a}}}\mathbf{=}\frac{\overline{){\mathbf{K}}_{\mathbf{a}}}\mathbf{·}{\mathbf{K}}_{\mathbf{b}}}{\overline{){\mathbf{K}}_{\mathbf{a}}}}\phantom{\rule{0ex}{0ex}}\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{a}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{b}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}}$

Kb = 5.556x10‒10

Kb is also an equilibrium expression and can be calculated as:

$\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}}$

Solids and liquids are ignored in equilibrium expressions

$\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}\frac{\left[{\mathbf{CH}}_{\mathbf{3}}\mathbf{COOH}\right]\left[{\mathbf{OH}}^{\mathbf{-}}\right]}{\left[{\mathbf{CH}}_{\mathbf{3}}{\mathbf{COO}}^{\mathbf{-}}\right]}}$

$\mathbf{5}\mathbf{.}\mathbf{556}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{=}\frac{\left[\mathbf{x}\right]\left[\mathbf{x}\right]}{\left[\mathbf{0}\mathbf{.}\mathbf{0217}\mathbf{-}\mathbf{x}\right]}$

###### Problem Details

Find the pH of the equivalence point(s) and the volume (mL) of 0.0372 M NaOH needed to reach the point(s) in titrations of

(a) 42.2 mL of 0.0520 M CH 3COOH