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lukranit [14]
3 years ago
13

A researcher wants to test the solubility (property of being dissolved) of salt in water as the temperature of the water increas

es.
Design an experiment to test his hypothesis that as the water temperature increases the solubility of the salt in water also increases. Include the following in your experimental design: experimental setup, procedure for data collection methods and criteria for evaluating the hypothesis.
Materials: salt, water, beakers, heat source, thermometer, balance
Chemistry
1 answer:
Leya [2.2K]3 years ago
3 0
The researcher may first weight the beaker with water and then start to heat the water to a constant temperature, for example 30 °C and then start adding salt and stirring. He should add salt slowly until solid salt starts to become visible and the solution starts becoming cloudy. When this happens, he should quickly weigh the beaker. The increase in mass is the mass of salt dissolved at that temperature.
The procedure is then repeated but at an increased temperature until 5-6 temperatures have been tested.
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How Many Moles Of HCl Need To Be Added To 150.0 ML Of 0.50 M NaZ To Have A Solution With A PH Of 6.50
Aleks04 [339]

The number of mole of HCl needed for the solution is 1.035×10¯³ mole

<h3>How to determine the pKa</h3>

We'll begin by calculating the pKa of the solution. This can be obtained as follow:

  • Equilibrium constant (Ka) = 2.3×10¯⁵
  • pKa =?

pKa = –Log Ka

pKa = –Log 2.3×10¯⁵

pKa = 4.64

<h3>How to determine the molarity of HCl </h3>
  • pKa = 4.64
  • pH = 6.5
  • Molarity of salt [NaZ] = 0.5 M
  • Molarity of HCl [HCl] =?

pH = pKa + Log[salt]/[acid]

6.5 = 4.64 + Log[0.5]/[HCl]

Collect like terms

6.5 – 4.64 = Log[0.5]/[HCl]

1.86 = Log[0.5]/[HCl]

Take the anti-log

0.5 / [HCl] = anti-log 1.86

0.5 / [HCl] = 72.44

Cross multiply

0.5 = [HCl] × 72.44

Divide both side by 72.44

[HCl] = 0.5 / 72.4

[HCl] = 0.0069 M

<h3>How to determine the mole of HCl </h3>
  • Molarity of HCl = 0.0069 M
  • Volume = 150 mL = 150 / 1000 = 0.15 L
  • Mole of HCl =?

Mole = Molarity x Volume

Mole of HCl = 0.0069 × 0.15

Mole of HCl = 1.035×10¯³ mole

<h3>Complete question</h3>

How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl

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