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denis-greek [22]

3 years ago

10

If 180 grams of potassium iodide is dissolved in 100 cm3 of water at 30oC, a(n) _______________ solution is formed.

2 answers:

olga nikolaevna [1]3 years ago

6 0

Super saturated solution is formed.

<u>Explanation:</u>

Solubility is the property of any substance's capacity, that is the solute of the substance is dissolved in the given solvent to form the solution. We have three different types of solution, unsaturated, saturated and supersaturated solution.

Unsaturated solution is a solution with lesser amount of solute than its solubility at equilibrium.

Saturated solution is a solution with the maximum solute dissolved in the solvent.

Super saturated solution is a solution with more solute than it is required.

The solubility of KI at 30°C is 153 g / 100 ml. Here 180 g of KI in 100 ml of water at 30°C is given, which has more solute than required, so it is super saturated solution.

hjlf3 years ago

4 0

Answer:

What they are trying to say is that its answer choice "D".

Explanation:

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3 years ago

A buffer solution contains 0.496 M hydrocyanic acid and 0.399 M sodium cyanide . If 0.0461 moles of sodium hydroxide are added t

pochemuha

Answer : The pH of the solution is, 9.63

Explanation : Given,

The dissociation constant for HCN = $pK_a=9.31$

First we have to calculate the moles of HCN and NaCN.

$\text{Moles of HCN}=\text{Concentration of HCN}\times \text{Volume of solution}=0.496M\times 0.225L=0.1116mole$

and,

$\text{Moles of NaCN}=\text{Concentration of NaCN}\times \text{Volume of solution}=0.399M\times 0.225L=0.08978mole$

The balanced chemical reaction is:

$HCN+NaOH\rightarrow NaCN+H_2O$

Initial moles 0.1116 0.0461 0.08978

At eqm. (0.1116-0.0461) 0 (0.08978+0.0461)

0.0655 0.1359

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$pH=9.31+\log (\frac{0.1359}{0.0655})$

$pH=9.63$

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Explanation:

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