The pH of the solution at 25 degree celsius of 1.3 × 10⁻⁶ moles of a sample of Sr(OH)₂ is 10.02.
<h3>How do we calculate pH?</h3>
The pH of any solution gives an idea about the acidic and basic nature of the solution and the equation of pH will be represented as:
pH + pOH = 14
Given that,
Moles of Sr(OH)₂ = 1.3 × 10⁻⁶ mol
Volume of solution = 25mL = 0.025L
The concentration of Sr(OH)₂ in terms of molarity = 1.3×10⁻⁶/0.025
= 5.2×10¯⁵M
Dissociation of Sr(OH)₂ takes place as:
Sr(OH)₂ → Sr²⁺ + 2OH⁻
From the stoichiometry of the reaction 1 mole of Sr(OH)₂ produces 2 moles of OH⁻.
Given that the base is a strong base and that it entirely dissociates into its ions, the hydroxide ion concentration is 5.2×10¯⁵×2 = 1.04×10¯⁴ M.
pOH = -log[OH⁻]
pOH = -log(1.04×10¯⁴)
pOH = 3.98
Now we put this value on the first equation we get,
pH = 14 - 3.98 = 10.02
Therefore, the value of pOH is 10.02.
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Write as a proportion, showing the relationship of both given information:
68.0g 0.3g
---------- = -----------
1L x ( your answer)
Cross multiply: 68.0g× X = 0.3g × 1L
68.0g (X)= 0.3g/L
Solve for X by dividing both sides by 68.0 g
68.0g (X) = 0.3g/L
------------- ------------------
68.0g 68.0g
Then enter into calculator 0.3/68 and that will be your solution. Make sure you round up.
Answer is A bc you can get electrocuted
Answer:
i also had this question:P
Explanation:
This is what i got The KB expression for aniline c6h5nh2 is: For C6H5NH2 + H2O >< C6H5NH3+ <span>OH-Kb = 4.3 x (10 ^ -10) = [C6H5NH3+][OH-] / [C6H5NH2]
hope this helps:)
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