75 mL5MHCl+25 mL5MNaOH 25 mL5MNaOH will neutralise 25 mL5MHCl 75−25=50 mL5MHCl will remain. Total volume will be 75+25=100 mL 50 mL5MHCl is diluted to 100 mL [H+]=[HCl]=5M×10050=10M pH=−log10[H+]=−log1010M=1
Following four solutions are prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to 1 ?
Held on 15 Apr 2018 · Verified 6 Jul 2026.
55 mL10MHCl+45 mL10MNaOH
75 mL5MHCl+25 mL5MNaOH
100 mL10MHCl+100 mL10MNaOH
60 mL10MHCl+40 mL10MNaOH
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