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MArishka [77]
3 years ago
8

Aqueous solutions of lithium hydroxide (LiOH) and acetic acid (HC2H3O2) will react to yield lithium acetate and water. In the ba

lanced equation for this reaction, what is the lowest possible whole number coefficient for water?
Chemistry
1 answer:
Nana76 [90]3 years ago
7 0

Answer: The lowest possible whole number coefficient for water is 1.

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble  water and remain in solid form are represented by (s) after their chemical formulas.

The balanced chemical reation for aqueous solutions of lithium hydroxide and acetic acid react to yield lithium acetate and water is:


LiOH(aq)+CH_3COOH(aq)\rightarrow CH_3COOLi(aq)+H_2O(l)

Thus the lowest possible whole number coefficient for water is 1.

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The amount of disorder in a system is measured by it's:
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4 years ago
A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is
Charra [1.4K]

Answer:

[H^{+}] = 0.761 \frac{mol}{L}

[OH^{-}]=1.33X10^{-14}\frac{mol}{L}

pH = 0.119

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L)  \\ n_{H^{+} } from HNO_{3}  = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

n_{H^{+} } from HCl = (5.00)(0.093)

n_{H^{+} } from HCl = 0.465 mol

n_{H^{+} } from HNO_{3}  = (8.00)(0.037)

n_{H^{+} } from HNO_{3}  = 0.296 mol

n_{H^{+}(total) } = 0.296 + 0.465

n_{H^{+}(total) } = 0.761 mol

For molar concentration of hydrogen ions:

[H^{+}]  = \frac{n_{H^{+}}(mol)}{V(L)}

[H^{+}] = \frac{0.761}{1.00}

[H^{+}] = 0.761 \frac{mol}{L}

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

K_{w} = [H^{+} ][OH^{-} ]

[OH^{-}]=\frac{Kw}{[H^{+}] }

[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }

[OH^{-}]=1.33X10^{-14}\frac{mol}{L}

The pH of the solution can be measured by the following formula:

pH = -log[H^{+} ]

pH = -log(0.761)

pH = 0.119

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