Question #12f80

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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

$Sxx84=0.7xx1$

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