Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations. (i) 60mL(M)/(10)HCl + 40mL(M)/(10)NaOH (ii) 55mL(M)/(10)HCl + 45mL(M)/(10)NaOH (iii) 75mL(M)/(5)HCl +

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Correct answer: D. (iii)

Equivalent of acid = N _ 1 V _ 1 Equivalent of base = N _ 2 V _ 2 As N _ 1 V _ 1 > N _ 2 V _ 2 , acid is left at the end of the reaction. The 'n' factor of NaOH and HCl are 1, 1 . So, N = M for both. N _ finalsolution = [ H ^ + ] = N _ 1 V _ 1 - N _ 2 V _ 2 V _ 1 + V _ 2 = ((1)/(5) × 75 - (1)/(5) × 25)/(75 + 25) = (1)/(10) = 10 ^ - 1

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