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9 months ago

What is the concentration of the hydroxide ion given that the concentration of the hydronium ion?

1 answer:

6 0

The concentration of the hydroxide ion given that the concentration of the hydronium ion 1.0 times 10^-14 M. The reverse mathematical method used to determine the pOH can be used to get the hydroxide ion concentration from the pOH. How many hydroxide ions are there in a solution with a pOH of 5.70, for instance.

Calculate 10-5.70, or "inverse" log, on a calculator (- 5.70). It indicates that one hydroxide ion is produced by one part of the NaOH solution. Because of this, the molar concentration of hydroxide ions in the solution is the same as the molar concentration of the NaOH.

To learn more about hydroxide, click here.

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2 years ago

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2 years ago

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2 years ago

Calculate the molarity of sodium ion in a solution made

Arada [10]

Answer:

0.1035 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Sodium chloride will furnish Sodium ions as:

$NaCl\rightarrow Na^{+}+Cl^-$

Given :

For Sodium chloride :

Molarity = 0.288 M

Volume = 3.58 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 3.58×10⁻³ L

Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:

$Moles =0.288 \times {3.58\times 10^{-3}}\ moles$

Moles of sodium ions by sodium chloride = 0.00103104 moles

Sodium sulfate will furnish Sodium ions as:

$Na_2SO_4\rightarrow 2Na^{+}+SO_4^{2-}$

Given :

For Sodium sulfate :

Molarity = 0.001 M

Volume = 6.51 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 6.51 ×10⁻³ L

Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:

$Moles =2\times 0.001 \times {6.51\times 10^{-3}}\ moles$

Moles of sodium ions by Sodium sulfate = 0.00001302 moles

Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles

Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L

Concentration of sodium ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{Na^+}=\frac{0.00104406}{10.09\times 10^{-3}}$

The final concentration of sodium anion = 0.1035 M

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3 years ago