A 25.0 mL sample of sulfuric acid is completely neutralized by 32.8 mL of 0.116 mol/L ammonia solution. Ammonium sulfate is formed What is the concentration of the sulfuric acid?

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A 25.0 mL sample of sulfuric acid is completely neutralized by 32.8 mL of 0.116 mol/L ammonia solution. Ammonium sulfate is formed What is the concentration of the sulfuric acid?

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Answer 1 · 2016-12-19T06:41:50

Approx. $0.08 \cdot m o l \cdot L^{- 1}$ $0.08*mol*L^-1$ with respect to sulfuric acid.

Explanation:

We need (i) a stroichiometric equation:

$H_{2} S O_{4} (a q) + 2 N H_{3} (a q) \to 2 N H_{4}^{+} + S O_{4}^{2 -}$ $H_2SO_4(aq) + 2NH_3(aq) rarr 2NH_4^+ + SO_4^(2-)$

At the 2nd equivalence point, the solution will stoichiometric in $ammonium sulfate$ $"ammonium sulfate"$ . An appropriate indicator must be used.

And (ii) a calculation,

$[H_{2} S O_{4}] = \frac{1}{2} \times \frac{32.8 \times 10^{- 3} \cdot L \times 0.116 \cdot m o l \cdot L^{- 1}}{25.0 \times 10^{- 3} \cdot L}$ $[H_2SO_4]=1/2xx(32.8xx10^-3*Lxx0.116*mol*L^-1)/(25.0xx10^-3*L)$

$= ? ? \cdot m o l \cdot L^{- 1}$ $=??*mol*L^-1$

Why did we use a $\frac{1}{2}$ $1/2$ factor in the calculation?

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A 25.0 mL sample of sulfuric acid is completely neutralized by 32.8 mL of 0.116 mol/L ammonia solution. Ammonium sulfate is formed What is the concentration of the sulfuric acid?

1 Answer

Explanation:

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