Question

A 1.235 g sample of H2S is dissolved in water and an acid-base indicator is added. The sample required 30.0 ml of 0.785 M NaOH to reach the equivalence point. What is the % by mass of H2S in the sample? To solve this problem, you would need the following: CHOOSE the best answer that answers all 3 parts (a, b, c) correctly.

Answer 1

The neutralization reaction involved is:
$H_{2}S(aq)\, +\, 2NaOH(aq)\, \rightarrow \, Na_{2}S(aq)\, +\, 2H_{2}O(l)$
At the equivalence point:
(G. Eq) of H2S = (G.Eq) of NaOH
Gram equivalents caan be calculated using any of the given formulas.
G.Eq = number of moles x n-factor
G.Eq = Molarity x Volume in L x n-factor
G.Eq = Normality x Volume in L
(G. Eq) of H2S = (G.Eq) of NaOH
(number of moles x n-factor ) of H2S = (Molarity x Volume in L x n-factor) of NaOH
Let the mass of H2S dissolved = w
(w/34 g x 2 ) of H2S = (0.785 x 30.0 x 10-3 x 1) of NaOH
w = 0.785 x 30.0 x 10-3 x 17 = 0.40035 g