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3 years ago

Calculate the ph of a 0.005 m solution of potassium oxide k2o

1 answer:

3 0

First, we have to see how K2O behaves when it is dissolved in water:

K2O + H20 = 2 KOH

According to reaction K2O has base properties, so it forms a hydroxide in water.
For the reaction next relation follows:

c(KOH) : c(K2O) = 1 : 2

So,

c(KOH)= 2 x c(K2O)= 2 x 0.005 = 0.01 M = c(OH⁻)

Now we can calculate pH:

pOH= -log c(OH⁻) = -log 0.01 = 2

pH= 14-2 = 12

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masha68 [24]

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2 years ago

All of the following methods would increase the solubility of a solid solute, except for

cricket20 [7]

May i please have a(n) answer choices please because it would be a lot better if it was like that and then ill answer it

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3 years ago

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noname [10]

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3 years ago

Read 2 more answers

A certain electrolyte solution contains 1 gram of salt for every 8 grams of sugar and every 200 grams of water. If the sugar to

12345 [234]

Answer:

The resulting solution contains approximately 666 g of water.

Explanation:

In the initial solution we have:

1g salt : 8g sugar : 200g water

This means that the ratios are:

$\frac{salt}{sugar} = \frac{1}{8} \\\\\frac{sugar}{water} = \frac{8}{200} =\frac{1}{25}$

In the final solution we have:

5g salt: xg sugar: yg water

The new ratios are:

$\frac{salt}{sugar} = \frac{3}{8} \\\\\frac{sugar}{water} = \frac{1}{50}$

Now we can calculate the amount of sugar in the final solution:

$\frac{salt}{sugar} = \frac{5}{x} =\frac{3}{8} \\\\X = 13.333 g$

Finally, we calculate the amount of water:

$\frac{sugar}{water} = \frac{13.333}{y} = \frac{1}{50} \\y = 666.667 g$

4 0

3 years ago

Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.

o-na [289]

Answer:

pKa = 7.46

Explanation:

1) Data:

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

2) Strategy:

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

3) Solution:

a) pH

pH = - log [H₃O⁺]

4.64 = - log [H₃O⁺]

$[H_3O^+]= 10^{-4.64} = 2.29.10^{-5}$

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

5 0

3 years ago