The ionization constant of an acid, Ka, is the measure of strength of an acid. The Ka values of acetic acid, hypochlorous acid and formic acid are 1.74 × 10−5, 3.0× 10−8and 1.8 × 10−4 respectively. Which of the following orders of pH of 0.1 mol dm-3 solutions of these acids is correct?
Answer : Option BExplaination / Solution: pH=-log(cα) , Ka=cα2 . Lesser is the pH value greater is acidity. Also greater is Ka, lesser will be acidity.
Q2.Ka1,Ka2 and Ka3 are the respective ionisation constants for the following reactions. H2S⇌H+ + HS−− HS−⇌H+ + S2− H2S⇌2H+ +S2− The correct relationship between Ka1,Ka2 and Ka3 is
Q6.Kafor CH3COOH is 1.8 × 10−5andKb for NH4OH is 1.8 × 10−5 The pH of ammonium acetate will be
Answer : Option AExplaination / Solution: CH3COONH4is a salt of weak acid and weak base. pH= 7+ 0.5(pKa-pKb) pKa=-log(Ka) and pKb=-log(Kb) Here pKa=pKb so pH=7.
On increasing the pressure, in which direction will the gas phase reaction proceed to re-establish equilibrium is predicted by applying the Le Chatelier’s principle. Consider the reaction. N2(g)+ 3H2(g) 2NH3 (g) Which of the following is correct, if the total pressure at which the equilibrium is established, is increased without changing the temperature?
Answer : Option DExplaination / Solution: Kc is independent of P change ,only the position at which equilibrium will be established will differ.
What will be the correct order of vapour pressure of water, acetone and ether At 30∘C. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point?
Answer : Option AExplaination / Solution:
Water has maximum boiling point due to hydrogen bonding so its vapour pressure will be least because of low boiling point less amount of water will move to vapour phase at a particulat temperature so lesser will be the vapour pressure.
Q10.At 500 K, equilibrium constant, Kc, for the following reaction is 5.2. 12H2(g)+12I2(g)⇌ HI (g) What would be the equilibrium constant Kc for the reaction 2HI (g) ⇌H2(g)+I2(g)
Answer : Option BExplaination / Solution: The second equation can be obtained by reversing the 1st reaction and multiplying by 2. So Kcnew=(1/Kcold)2 Here ,Kcold=5.2
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