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Mazyrski [523]
3 years ago
5

What is the pH of a solution with [H+]=6.2x10-9 M?

Chemistry
1 answer:
Serga [27]3 years ago
3 0

Answer: 8.2

Explanation:

pH of a solution is  - Log [ H+].

pH = - Log [ 6.2 x 10-9 M]

     = 9 - 0.7924

pH =  8.24 approx 8.2

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A certain liquid X has a normal freezing point of 7.60 °C and a freezing point depression constant K= 6.90 °C-kg-mol. Calculate
Dmitry [639]

<u>Answer:</u> The freezing point of solution is -5.11°C

<u>Explanation:</u>

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of sodium chloride follows:

NaCl(aq.)\rightarrow Na^{+}(aq.)+Cl^-(aq.)

The total number of ions present in the solution are 2.

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (NaCl) = 7.57 g

M_{solute} = Molar mass of solute (NaCl) = 58.44 g/mol

W_{solvent} = Mass of solvent (liquid X) = 350.0 g

Putting values in above equation, we get:

\text{Molality of }NaCl=\frac{7.57\times 1000}{58.44\times 350.0}\\\\\text{Molality of }NaCl=0.370m

To calculate the depression in freezing point, we use the equation:

\Delta T=iK_fm

where,

i = Vant hoff factor = 2

K_f = molal freezing point depression constant = 6.90°C/m

m = molality of solution = 0.370 m

Putting values in above equation, we get:

\Delta T=2\times 6.90^oC/m.g\times 0.370m\\\\\Delta T=5.11^oC

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T=\text{freezing point of water}-\text{freezing point of solution}

\Delta T = 5.11 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

5.106^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-5.11^oC

Hence, the freezing point of solution is -5.11°C

7 0
4 years ago
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