A soft drink was bottled with a partial pressure of CO2
of 3 bar over the liquid at room temperature.
The partial pressure of CO2
over the solution approaches a value of 30 bar when 44 g of CO2
is
dissolved in 1 kg of water at room temperature. The approximate pH of the soft drink is ______ $\times$ 10–1.
(First dissociation constant of
H2CO3
= 4.0 $\times$ 10–7; log 2 = 0.3; density
of the soft drink = 1 g mL–1)
.
Answer (integer)
37
Solution
CO<sub>2</sub>
+ H<sub>2</sub>O $\to$ H<sub>2</sub>CO<sub>3</sub>
<br><br>At 30 bar pressure mass of CO<sub>2</sub> in 1 kg water
= 44 gm
<br><br>At 3 bar pressure mass of CO<sub>2</sub> in 1 kg water
= 4.4 gm
<br><br>$\therefore$ Moles of CO<sub>2</sub> in 1 kg water = ${{4.4} \over {44}}$ = 0.1
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<table class="tg">
<thead>
<tr>
<th class="tg-baqh"></th>
<th class="tg-baqh">H<sub>2</sub>CO<sub>3</sub></th>
<th class="tg-baqh">⇌</th>
<th class="tg-baqh">H<sup>+</sup></th>
<th class="tg-baqh">+</th>
<th class="tg-baqh">HCO<sub>3</sub><sup>-</sup></th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">t = 0</td>
<td class="tg-baqh">0.1</td>
<td class="tg-baqh"></td>
<td class="tg-baqh">0</td>
<td class="tg-baqh"></td>
<td class="tg-baqh">0</td>
</tr>
<tr>
<td class="tg-baqh">t = t<sub>eq</sub></td>
<td class="tg-baqh">0.1(1 - $\alpha$)</td>
<td class="tg-baqh"></td>
<td class="tg-baqh">0.1$\alpha$</td>
<td class="tg-baqh"></td>
<td class="tg-baqh">0.1$\alpha$</td>
</tr>
</tbody>
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<br><br>4.0 $\times$ 10<sup>–7</sup> = ${{0.1{\alpha ^2}} \over {1 - \alpha }}$
<br><br>${1 - \alpha }$ $\simeq$ 1
<br><br>$\Rightarrow$ 0.1${{\alpha ^2}}$ = 4 $\times$ 10<sup>-7</sup>
<br><br>$\Rightarrow$ $\alpha$ = 2 $\times$ 10<sup>-3</sup>
<br><br>[H<sup>+</sup>] = 0.1$\alpha$ = 2 $\times$ 10<sup>-4</sup>
<br><br>$\therefore$ pH = –[– 4 × log(2)] = 3.7 = 37 × 10<sup>–1</sup>
About this question
Subject: Chemistry · Chapter: Equilibrium · Topic: Chemical Equilibrium and Kc, Kp
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