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Answer 1 · 2016-10-20T13:55:51

Molar mass of NaOH #=(23+16+1)=40" g/mol"#

Equivalent mass of NaOH#="Molar mass"/"Acidity"=40" g/equivalent"#

Strength of 0.8% NaOH#=8" g/L"=(8g/L)/(40g/"equivalent")#

#=0.2(N)#

Acid taken 25 mL 0.5(N) HCl#equiv12.5mL" " 1 (N) HCl#

Base taken 59mL 0.2(N) NaOH

#equiv 11.8mL" "1(N) NaOH #

After partial neutralisation the remaining HCl will be
#equiv(12.5-11.8)=0.7mL" "1 (N) HCl#

Total volume of resulting solution #(25+59)=84mL#

If the strength of this resulting HCl solution be #S(N)# then

#Sxx84=0.7xx1#

#S=0.7/84=0.0083(N)#