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3 years ago

In the soap testing experiment, what is the purpose of adding some MgCl2 to the soap solution at the end?

Chemistry

1 answer:

lakkis [162]3 years ago

4 0

Explanation:

The water is hard due to the presence of ions of Mg²⁺. Now in hard water soaps are ineffective . Since in hard water Mg²⁺ ion forms precipitate , which concludes less number of soap molecules are present in the solution and less amount of froth which disables the cleansing property of the soap .

Now the experiment , the very last step is the hard water test where some amount of MgCl₂ is added ,

Now this magnesium salt act as a source of Mg²⁺ ion and now the soap action can be determined , whether it is able to form froth or not .

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Answer:

[H₃O⁺] = 1.4 × 10⁻⁹ M.

Explanation:

NH₄Cl is a salt that dissolves well in water. The 2.5 M NH₄Cl will give an initial NH₄⁺ concentration of 2.5 M.

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Let the increase in OH⁻ concentration be x. The initial OH⁻ concentration at room temperature is 10⁻⁷ M.

Construct a RICE table for the equilibrium between NH₃ and NH₄⁺:

$\begin{array}{c|ccccccc}\text{R}&\text{NH}_3 &+&\text{H}_2\text{O}&\rightleftharpoons &{\text{NH}_4}^{+}&+&\text{OH}^{-}\\\text{I}&1.0&&&&2.5&&1.0\times 10^{-7}\\\text{C}& -x &&&& +x &&+x\\\text{E} &1.0 - x &&&&2.5+x&&1.0\times 10^{-7}+x\end{array}$ .

The $\text{K}_b$ value for ammonia is small. The value of x will be so small that at equilibrium, $1.0 - x \approx 1.0$ and $2.5- x \approx 2.5$ .

$\displaystyle \text{K}_b = \frac{[{\text{NH}_4}^{+}]\cdot [{\text{OH}}^{-}]}{[\text{NH}_3]} \approx \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0}$ .

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$\displaystyle [\text{OH}^{-}] = x+1.0\times 10^{-7} = 1.8\times 10^{-5} /\left(\frac{2.5}{1.0}\right) = 7.2\times 10^{-6}\;\text{mol}\cdot\text{L}^{-}$ .

Again, $\text{K}_w = 1.0\times 10^{-14}$ at room temperature.

$\displaystyle [\text{H}_3\text{O}^{+}] = \frac{\text{K}_w}{[\text{OH}^{-}]}=\frac{1.0\times 10^{-14}}{7.2\times 10^{-6}} = 1.4\times 10^{-9} \;\text{mol}\cdot\text{L}^{-1}$

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