The pH of the diluted HCl solution is 1.3.
Explanation:
Given:
The concentrated HCl solution of 8.0 M. The 1.5 mL of 8.0 M HCl is diluted with water to 250 mL volume.
To find:
The pH of the diluted HCl solution.
Solution
- The concentration of the HCl solution before dilution =

- The volume of the HCl solution taken for dilution =

- The concentration of the HCl solution after dilution =

- The volume of the HCl solution after dilution =

Using the Dilution equation:

The concentration of diluted HCl solution = 0.048 M

In the 1 M solution of HCl, there are 1 M of hydrogen ion, then the concentration of hydrogen ions in 0.048 M of HCl will be:
![[H^+]=1\times 0.048M=0.048 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1%5Ctimes%200.048M%3D0.048%20M)
The pH of the diluted HCl solution :
![pH=-\log [H^+]\\=-\log [0.048M]=1.18 \approx 1.3](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D%5C%5C%3D-%5Clog%20%5B0.048M%5D%3D1.18%20%5Capprox%201.3)
The pH of the diluted HCl solution is 1.3.
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Answer:
It will exert the pressure at a temperature of <em><u>153.44 K</u></em>
Explanation:
To answer this question, we shall be using the ideal gas equation;
PV = nRT
Since we are calculating the temperature, it can be made the subject of the formula.
Thus, this can be T = PV/nR
where P is the pressure = 900 torr
V is the volume = 0.75 L
n is the number of moles = 0.0705 mol
R is the molar gas constant = 62.4 L.Torr.
.
Plugging these values into the equation, we have;
T = (900 × 0.75)/(0.0705 × 62.4)
T = 153.44 K
The pressure, in mmHg, exerted by only the oxygen gas is 720.8 mmHg
<h3>Pressure of gas collected over water</h3>
The total pressure of a gas collected over water is given by P = P' + p where
- P = pressure of oxygen gas and
- p = vapour pressure of water at 25 °C = 25 mmHg
Making P' subject of the formula, we have
P' = P - p
Since the total pressure of oxygen plus water vapor is P = 745.8 mmHg, substituting the values of the variables into the equation, we have
P' = P - p
P' = 745.8 mmHg - 25 mmHg
P' = 720.8 mmHg
So, the pressure, in mmHg, exerted by only the oxygen gas is 720.8 mmHg
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Answer:
particles
Explanation:
hope these help and have a blessed day