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3 years ago

The kw for water at 0 °c is 0.12× 10–14 m2. Calculate the ph of a neutral aqueous solution at 0 °c.

Chemistry

1 answer:

jekas [21]3 years ago

8 0

Answer:

pH = 7.46.

Explanation:

The ionization of water is given by the equation :

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq),

The equilibrium constant (Kw) expression is:

Kw = [H⁺][OH⁻] = 0.12 x 10⁻¹⁴.

in pure water and neutral aqueous solution, [H⁺] = [OH⁻]

So, Kw = [H⁺]²

∴ 0.12 x 10⁻¹⁴ = [H⁺]²

∴ [H⁺] = 3.4 x 10⁻⁸ M.

∵ pH = - log [H⁺]

pH = - log (3.4 x 10⁻⁸) = 7.46.

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3 years ago

Following the instructions in your lab manual, you have titrated a 25.00 mL sample of 0.0100 M KIO3 with a solution of Na2S2O3 o

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For 1: The amount of potassium iodate that were titrated is $2.5\times 10^{-4}$ moles

For 2: The amount of sodium thiosulfate required is $1.25\times 10^{-4}$ moles

Explanation:

For 1:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

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Putting values in above equation, we get:

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Hence, the amount of potassium iodate that were titrated is $2.5\times 10^{-4}$ moles

For 2:

The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:

$2KIO_3+Na_2S_2O_3\rightarrow K_2S_2O_3+2NaIO_3$

By Stoichiometry of the reaction:

2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate

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