Question #b8733

Question

1430 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-24T05:45:30

In case of acid base neutralization we know the relation

$S_{a} \times V_{a} = S_{b} \times V_{b}$ $S_axxV_a=S_bxxV_b$

Where

$S_{a} \to strength of acid solution in Normality = ?$ $S_a-> "strength of acid solution in Normality" =?$

$V_{a} \to Volume of acid solution = 2 m L$ $V_a->"Volume of acid solution"=2mL$

$S_{b} \to strength of base solution in Normality = \frac{N}{5}$ $S_b-> "strength of base solution in Normality"=N/5$

$V_{b} \to Volume of acid solution = 10 m L$ $V_b->"Volume of acid solution"=10mL$

Hence

$S_{a} \times V_{a} = S_{b} \times V_{b}$ $S_axxV_a=S_bxxV_b$

$\Rightarrow S_{a} \times 2 = \frac{N}{5} \times 10$ $=>S_axx2=N/5xx10$

$\Rightarrow S_{a} = 1 (N)$ $=>S_a=1(N)$