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Question
All solution are at 25∘C
Column I Column II(I) A solution containing weak acid 0.1 M HA (Ka=10−5& 0.1M NaA)(P) Buffer solution(II) In a mixture of 200 ml 0.5 M Na3PO4 and 400 ml 0.5 M HCl(Q) pH = 10 (for H3PO4K1=10−3,K2=10−7,K3=10−12)(III) An aqueous solution of sparingly soluble A(OH)3(Ksp=27×10−40)(R) Acidic solution(IV) 0.1 MNa2A (Ka1(H2A)=10−3),Ka2(H2A)=10−7solution(S) pH is closer to 7(T) pH = 5
Which of the following options has the correct combination considering LIst - I and List - II?
All solution are at 25∘C
Column I Column II(I) A solution containing weak acid 0.1 M HA (Ka=10−5& 0.1M NaA)(P) Buffer solution(II) In a mixture of 200 ml 0.5 M Na3PO4 and 400 ml 0.5 M HCl(Q) pH = 10 (for H3PO4K1=10−3,K2=10−7,K3=10−12)(III) An aqueous solution of sparingly soluble A(OH)3(Ksp=27×10−40)(R) Acidic solution(IV) 0.1 MNa2A (Ka1(H2A)=10−3),Ka2(H2A)=10−7solution(S) pH is closer to 7(T) pH = 5
Which of the following options has the correct combination considering LIst - I and List - II?
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Solution
The correct option is C (IV),(Q)
(I)
Buffer solution
pKa=−logKapKa=−log10−5pKa=5pH=pKa+logsaltacidpH=5+log0.10.1pH=5+log1pH=5∴pH=−log[H+]H+=10−5
Hence, for (I) - P,R,T are correct
(II)
200 mL 0.5MNa3PO4 and 400 mL 0.5 M HCl
Na3PO4 + HCl → NaH2PO4200mL×0.5M 400mL×0.5M100 mole 200 mole 100 moleAs NaH2PO4 is amphoteric salt, so pH can be calculated as
pH=pKa1+pKa22pH=3+72
pH=5.0
For (II), R,T are correct.
(III)
A(OH)3→A++3OH−s3s
Ksp=s.(3s)3Ksp=27s4
Given, Ksp=27×10−40
Therefore s=10−10OH−=3×10−10OH−net=OH−H2O+OH−A(OH)3OH−net=OH−H2OOH−=10−7pH=−log[H+]pH=−log[10−7]pH=7
For (III) only S is correct.
(IV)
A2−+H2O→HA++OH−h=√KwKa2×C (∴√Kh×C Kh=10−7)=√10−7×0.1=10−4=[OH−][H+]=10−1410−4=10−10pH=10
For IV only Q is correct. Hence, option c is the correct answer.
(I)
Buffer solution
pKa=−logKapKa=−log10−5pKa=5pH=pKa+logsaltacidpH=5+log0.10.1pH=5+log1pH=5∴pH=−log[H+]H+=10−5
Hence, for (I) - P,R,T are correct
(II)
200 mL 0.5MNa3PO4 and 400 mL 0.5 M HCl
Na3PO4 + HCl → NaH2PO4200mL×0.5M 400mL×0.5M100 mole 200 mole 100 moleAs NaH2PO4 is amphoteric salt, so pH can be calculated as
pH=pKa1+pKa22pH=3+72
pH=5.0
For (II), R,T are correct.
(III)
A(OH)3→A++3OH−s3s
Ksp=s.(3s)3Ksp=27s4
Given, Ksp=27×10−40
Therefore s=10−10OH−=3×10−10OH−net=OH−H2O+OH−A(OH)3OH−net=OH−H2OOH−=10−7pH=−log[H+]pH=−log[10−7]pH=7
For (III) only S is correct.
(IV)
A2−+H2O→HA++OH−h=√KwKa2×C (∴√Kh×C Kh=10−7)=√10−7×0.1=10−4=[OH−][H+]=10−1410−4=10−10pH=10
For IV only Q is correct. Hence, option c is the correct answer.
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