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Blababa [14]
3 years ago
8

Calculate the pH of each solution at 25∘C

Chemistry
1 answer:
JulijaS [17]3 years ago
8 0

Answer: pH of HCl =5, HNO3 = 1,

NaOH = 9, KOH = 12

Explanation:

pH = -log [H+ ]

1. 1.0 x 10^-5 M HCl

pH = - log (1.0 x 10^-5)

= 5 - log 1 = 5

2. 0.1 M HNO3

pH = - log (1.0 x 10 ^ -1)

pH = 1 - log 1 = 1

3. 1.0 x 10^-5 NaOH

pOH = - log (1.0 x 10^-5)

pOH = 5 - log 1 = 5

pH + pOH = 14

Therefore , pH = 14 - 5 = 9

4. 0.01 M KOH

pOH = - log ( 1.0 x 10^ -2)

= 2 - log 1 = 2

pH + pOH = 14

Therefore, pH = 14 - 2 = 12

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Answer:

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Explanation:

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3 years ago
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A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being
Igoryamba

Answer:

<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>

Explanation:

This question is about solubility.

Regarding solubility, the solutions may be classified as:

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  • Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.

Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.

  • In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.

  • In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g  of water.

  • Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ

       115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O  ⇒ x =  57.5 g NaNO₃

  • <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>

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3 years ago
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Using charles law
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v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2  cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius
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Ionic would be the answer
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Answer:

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