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3 years ago

5

To standardize a hydrochloric acid solution, it was used as a titrant with a solid sample of sodium hydrogen carbonate, NaHCO3.

The solid sample had a mass of 0.3967g, and 41.77 mL of acid was required to reach the equivalence point. Calculate the concentration of the standard solution.

1 answer:

Ket [755]3 years ago

6 0

Answer:

0.113 M

Explanation:

The reaction that takes place is:

NaHCO₃ + HCl →NaCl + CO₂ + H₂O

First we convert 0.3967 g of NaHCO₃ into moles, using its molar mass:

0.3967 g ÷ 84 g/mol = 4.72x10⁻³ mol NaHCO₃

As 1 mol of NaHCO₃ reacts with 1 mol of HCl, in 41.77 mL of the HCl solution there were 4.72x10⁻³ moles of HCl.

With the <em>calculated number of moles and the given volume </em>we <u>calculate the concentration of the solution</u>:

Converting 41.77 mL ⇒ 41.77 mL / 1000 = 0.04177 L

Concentration = 4.72x10⁻³ mol / 0.04177 L = 0.113 M

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$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

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Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

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<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

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and,

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$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

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<u>Given:</u>

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

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