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Answer 1 · 2016-10-20T13:55:51

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

Equivalent mass of NaOH $= \frac{Molar mass}{Acidity} = 40 g/equivalent$ $="Molar mass"/"Acidity"=40" g/equivalent"$

Strength of 0.8% NaOH $= 8 g/L = \frac{8 \frac{g}{L}}{40 \frac{g}{equivalent}}$ $=8" g/L"=(8g/L)/(40g/"equivalent")$

$= 0.2 (N)$ $=0.2(N)$

Acid taken 25 mL 0.5(N) HCl $\equiv 12.5 m L 1 (N) H C l$ $equiv12.5mL" " 1 (N) HCl$

Base taken 59mL 0.2(N) NaOH

$\equiv 11.8 m L 1 (N) N a O H$ $equiv 11.8mL" "1(N) NaOH$

After partial neutralisation the remaining HCl will be
$\equiv (12.5 - 11.8) = 0.7 m L 1 (N) H C l$ $equiv(12.5-11.8)=0.7mL" "1 (N) HCl$

Total volume of resulting solution $(25 + 59) = 84 m L$ $(25+59)=84mL$

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

$S \times 84 = 0.7 \times 1$ $Sxx84=0.7xx1$

$S = \frac{0.7}{84} = 0.0083 (N)$ $S=0.7/84=0.0083(N)$