Q.

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 44g  of CO2  is dissolved in 1 kg of water at room temperature. The approximate pH of the soft drink is ______×10−1(First dissociation constant of H2CO3=4.0×10−7;log2=0.3; density of the soft drink =1 g  mL−1)

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answer is 37.

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Detailed Solution

Herny’s law : - ωg∝P(for 1 kg of ssolvent)In 100rg  of  H2Oωg1ωg2=P1P2 ⇒Δ4ω=303⇒ω=4.4g of ω2=0.1 moles of CO2 Wt. of solution =1000+4.4=1004.4g∴Vsol =1004.41=1004.4ml=1.0044l∴CO2=0.11.0044≃0.3M=H2CO3H2CO3⇌KalH++HCO3−H+=Ka1×C=4×10−7×10−1=2×10−4PH= --log (H+)
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