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Q.

A cell diagram shown below contains one litre of buffer solution of HA(pKa = 4) and NaA in left side compartment and same weak acid (HA) solution in right side. Then the value of Ecell  is (y×10−2V). The value of ‘y’ is  [2.303RTF=0.06]

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answer is 6.

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Detailed Solution

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In Anode half cell,  [H+]LHE=Ka[Acid][salt]=10−4×10.1=10−3M
In Cathode half cell,  [H+]RHE=KaC=10−4×1=10−2M
Ecell=−0.061log10−310−2=0.06=6×10−2V

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