Question

200ml decimolar H2SO4 and 300ml decimolar HCl mixture is treated with M/5 NaOH solution. For complete neutralization of the mixture, the volume of NaOH needed is
(1) 400ml (2) 250ml (3) 350ml (4) 100ml

Answer 1

Answer:

350

Explanation:

No. of equivalents of acid=no. of equivalents of base

$200 \times 0.2 + 300 \times 0.1 = \frac{1}{5} \times v \\ \frac{v}{5} = 40 + 30 = 70 \\ v = 350$

Answer 2

Answer:

350

Explanation:

No. of equivalents of acid=no. of equivalents of base

\begin{gathered}200 \times 0.2 + 300 \times 0.1 = \frac{1}{5} \times v \\ \frac{v}{5} = 40 + 30 = 70 \\ v = 350\end{gathered}

200×0.2+300×0.1=

5

1

×v

5

v

=40+30=70

v=350