hat will be the osmotic pressure of 0.02M monoacidic base if it's pH is 12 at 300 K? (1) 0.74 atm (2) 0.37 atm (3) 3.6 atm (4) None of the above
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First of all we have to find out concentration of [H⁺]
pH = -log[H⁺]
12 = -log[H⁺]
[H⁺] = 10⁻¹² M
∵[H⁺][OH⁻] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴/10⁻¹² = 10⁻² = 0.01M
Now, use [OH⁻] = Cα
Given, C = 0.02
so, α = 0.01/0.02 = 0.5
Now, use van Hoff's factor
i = α + 1
= 0.5 + 1 = 1.5
∴ osmotic pressure , π = iCRT
= 1.5 × 0.02 RT
= 0.03RT
Now, put R = 0.082 and T = 300K
π = 0.03 × 0.082 × 300
= 9 × 0.082 atm
= 0.738atm ≈ 0.74 atm
Hence, option (1) is correct.
pH = -log[H⁺]
12 = -log[H⁺]
[H⁺] = 10⁻¹² M
∵[H⁺][OH⁻] = 10⁻¹⁴
[OH⁻] = 10⁻¹⁴/10⁻¹² = 10⁻² = 0.01M
Now, use [OH⁻] = Cα
Given, C = 0.02
so, α = 0.01/0.02 = 0.5
Now, use van Hoff's factor
i = α + 1
= 0.5 + 1 = 1.5
∴ osmotic pressure , π = iCRT
= 1.5 × 0.02 RT
= 0.03RT
Now, put R = 0.082 and T = 300K
π = 0.03 × 0.082 × 300
= 9 × 0.082 atm
= 0.738atm ≈ 0.74 atm
Hence, option (1) is correct.
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