3 g of acetic acid is added to 250 mL of 0.1 M HCL and the solution made up to 500 mL. To 20 mL
of this solutions ${1 \over 2}$ mL of 5 M NaOH is added. The pH of the solution is __________.
[Given : pKa of acetic acid = 4.75, molar mass of acetic of acid = 60 g/mol, log 3 = 0.4771]
Neglect any changes in volume.
Answer (integer)
5
Solution
milimole of acetic acid in 20 ml
<br>= ${3 \over {60}} \times 1000 \times {{20} \over {500}}$ = 2
<br><br>milimole of HCl in 20 ml = 25 $\times$ ${{20} \over {500}}$ = 1
<br><br>milimole of NaOH 20 ml = ${1 \over 2} \times 5$ = 2.5
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<th class="tg-9wq8">NaOH</th>
<th class="tg-9wq8">+</th>
<th class="tg-9wq8">CH<sub>3</sub>COOH</th>
<th class="tg-9wq8">$\to$</th>
<th class="tg-9wq8">CH3COONa</th>
<th class="tg-baqh">+</th>
<th class="tg-baqh">H<sub>2</sub>O</th>
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<td class="tg-9wq8"></td>
<td class="tg-9wq8">1.5</td>
<td class="tg-9wq8">2</td>
<td class="tg-9wq8"></td>
<td class="tg-9wq8">0</td>
<td class="tg-baqh"></td>
<td class="tg-baqh">0</td>
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<td class="tg-baqh"></td>
<td class="tg-baqh">0</td>
<td class="tg-baqh">0.5</td>
<td class="tg-baqh"></td>
<td class="tg-baqh">1.5</td>
<td class="tg-baqh"></td>
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<br><br>pH = pKa + log${{1.5} \over {0.5}}$
<br><br>= 4.74 + log 3 = 5.22
About this question
Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp
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