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FinnZ [79.3K]
3 years ago
11

What are the concentrations of hydroxide and hydronium ions in a solution with a pH of 8.3?

Chemistry
1 answer:
Margarita [4]3 years ago
5 0

Answer:

The hydroxide concentration is 2*10^-6 M

The hydronium concentration is 5*10^-9 M

Explanation:

Step 1: Data given

pH = 8.3

Step 2: Calculate [H+]

pH = 8.3 = -log [H+]

[H+] = 10^-8.3 = 5*10^-9 M

[H+] = [H3O+] = 5*10^-9 M

Step 3: Calculate pOH

pH + pOH = 14

pOH = 14 -pH = 14 - 8.3 = 5.7

Step 4: Calculate concentration of hydroxide

pOH = -log [OH-] = 5.7

[OH-] = 10^-5.7 = 2*10^-6 M

The hydroxide concentration is 2*10^-6 M

The hydronium concentration is 5*10^-9 M

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When two atoms with different electronegativities are bonded together, a bond dipole exists. These are displayed with an arrow o
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Answer:

Dipole, less electronegativity,  higher electronegativity

Explanation:

!!! Incomplete Question

When two atoms with different electronegativities are bonded together, a bond dipole exists.

A bond dipole is a partial charge assigned to bonded atoms due to difference in electron density, difference in electronegativity is a factor to this.

These are displayed with an arrow originating at the atom with the less electronegativity and pointing toward the atom with the higher electronegativity value

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3 years ago
A sample of 9.27 g9.27 g of solid calcium hydroxide is added to 38.5 mL38.5 mL of 0.500 M0.500 M aqueous hydrochloric acid. Writ
navik [9.2K]

<u>Answer:</u> The excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.  

<u>Explanation:</u>

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}   ....(1)

<u>For calcium hydroxide:</u>

Given mass of calcium hydroxide = 9.27 g

Molar mass of calcium hydroxide = 74.093 g/mol

Putting values in above equation, we get:

\text{Moles of calcium hydroxide}=\frac{9.27g}{74.093g/mol}=0.125mol

  • To calculate the moles of a solute, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

We are given:

Volume of hydrochloric acid = 38.5mL = 0.0385 L   (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.500 moles/ L

Putting values in above equation, we get:

0.500mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0385L}\\\\\text{Moles of hydrochloric acid}=0.01925mol

  • For the given chemical equation:

2HCl(aq.)+Ca(OH)_2(s)\rightarrow CaCl_2(s)+2H_2O(l)

Here, the solid salt is calcium chloride.

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.

So, 0.01925 moles of hydrochloric acid will react with = \frac{1}{2}\times 0.01925=0.009625moles of calcium hydroxide.

As, given amount of calcium hydroxide is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

  • Amount of excess reagent (calcium hydroxide) left = 0.125 - 0.01925 = 0.115375 moles

By Stoichiometry of the reaction:

2 moles of hydrochloric acid produces 1 mole of calcium chloride.

So, 0.01925 moles of hydrochloric acid will produce = \frac{1}{2}\times 0.01925=0.009625moles of calcium chloride.

Now, calculating the mass of calcium chloride from equation 1, we get:

Molar mass of calcium chloride = 110.98 g/mol

Moles of calcium chloride = 0.009625 moles

Putting values in equation 1, we get:

0.009625mol=\frac{\text{Mass of calcium chloride}}{110.98g/mol}\\\\\text{Mass of calcium chloride}=1.068g

Hence, the excess reagent for the given chemical reaction is calcium hydroxide and the amount left after the completion of reaction is 0.115375 moles. The amount of calcium chloride formed in the reaction is 1.068 grams.

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