Write an equation showing how this buffer neutralizes added base (csoh).

1 answer:

5 0

Here, the buffer is NH₄Cl + NH₃
The strong base is added to the buffer solution, then the salt reacts with base to form weak base and water, which does not alter the pH

NH₄Cl + CsOH → NH₃ + CsCl + H₂O
NH₄⁺ + OH⁻ → NH₃ + H₂O

When we add a strong acid to the buffer solution, the base NH₃ reacts with acid and forms salt, which does not alter the pH
NH₃ + H⁺ → NH₄⁺

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<h3>Explanation</h3>

Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as $\text{Pb}^{2+}$ ions. The formula for a nitrate ion is ${\text{NO}_3}^{-}$ . The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each $\text{Pb}^{2+}$ ion in lead nitrate would pair up with two ${\text{NO}_3}^{-}$ ions. The formula for lead nitrate will be $\text{Pb}({\text{NO}_3})_2$ . Each formula unit of lead nitrate will contain one $\text{Pb}^{2+}$ ion and two ${\text{NO}_3}^{-}$ ions.

The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as $\text{Cr}^{3+}$ . Similarly, each $\text{Cr}^{3+}$ will pair up with three ${\text{NO}_3}^{-}$ ions. The formula for chromium (III) nitrate will be $\text{Cr}(\text{NO}_3})_3$ . Each formula unit of chromium (III) nitrate will contain one ${\text{NO}_3}^{-}$ ion and three ${\text{NO}_3}^{-}$ ions.

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