Answer to Question #101131 in General Chemistry for Progress

Question

A is 0.125mol/dm^3 of H2S04. B is a solution containing xg/dm^3 of NaOH. The volume of the base is 25cm^3. At the end of the titration experiment, the volume of the acidic used was 24.5cm^3

Answer 1

Solution.

$C(H2SO4) \times V(H2SO4) = C(NaOH) \times V(NaOH)$

C(H2SO4) = 0.125 * 2 = 0.25 mol-eq/dm3

$C(NaOH) = \frac{0.25 \times 0.0245}{0.025} = 0.245 \ \frac{mol-eq}{dm^3}$

C(NaOH) = 0.245 mol/dm3

Because the alkali equivalence factor is 1.

$C'(NaOH) = C(NaOH) \times M(NaOH)$

C'(NaOH) = 9.80 g/dm3

Answer:

C'(NaOH) = 9.80 g/dm3