The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Solubility of Sr(OH)2 = 19.23 g/L
Then, concentration of Sr(OH)2
= 19.23 / 121.63 M
= 0.1581 M
Sr(OH)2(aq) → Sr2+(aq) + 2 (OH-)(aq)
∴Sr2+ = 0.1581M
[OH-] = 2 x 0.1581M = 0.3126 M
Now
Kw = [OH-] [H+]
10-14 / 0.3126 = [H+]
⇒ [H+] = 3.2 x 10-14
∴ pH = 13.495 = 13.50
Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:
Isotope |
Isotopic molar mass |
Abundance |
36Ar |
35.96755 gmol–1 |
0.337% |
38Ar |
37.96272 gmol–1 |
0.063% |
40Ar |
39.9624 gmol–1 |
99.600% |
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 58: The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of stronti....
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