Consider the following statements:(a) The pH | vedclass

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(B) $(a) \, H_2SO_4 + 2NaOH 	o Na_2SO_4 + 2H_2O$Initial millimoles of $H_2SO_4 = 400 	imes 0.1 = 40 \, mmol$.Initial millimoles of $NaOH = 400 	ime

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$(a), (b)$ and $(c)$

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(B) $(a) \, H_2SO_4 + 2NaOH 	o Na_2SO_4 + 2H_2O$Initial millimoles of $H_2SO_4 = 400 	imes 0.1 = 40 \, mmol$.Initial millimoles of $NaOH = 400 	imes 0.1 = 40 \, mmol$.Since $1 \, mol \, H_2SO_4$ reacts with $2 \, mol \, NaOH$,$20 \, mmol$ of $H_2SO_4$ will react with $40 \, mmol$ of $NaOH$.Remaining $H_2SO_4 = 40 - 20 = 20 \, mmol$.Total volume $= 800 \, mL$.$[H^{+}] = \frac{20 	imes 2}{800} = \frac{40}{800} = 0.05 \, M = 5 	imes 10^{-2} \, M$.$pH = -\log(5 	imes 10^{-2}) = 2 - \log(5) = 2 - 0.699 \approx 1.3$. Thus,$(a)$ is correct.$(b)$ The ionic product of water $(K_w)$ is an equilibrium constant,which is temperature dependent. Thus,$(b)$ is correct.$(c)$ For a monobasic acid,$K_a = \frac{c\alpha^2}{1-\alpha} = \frac{[H^{+}]\alpha}{1-\alpha}$. Given $K_a = 10^{-5}$ and $pH = 5$ $([H^{+}] = 10^{-5})$,we have $10^{-5} = \frac{10^{-5} \cdot \alpha}{1-\alpha}$,which gives $1-\alpha = \alpha$,so $\alpha = 0.5$ or $50 \%$. Thus,$(c)$ is correct.$(d)$ The common-ion effect is a direct application of Le Chatelier's principle. Thus,$(d)$ is incorrect.

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